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• \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi
• \cos (x)-\sin (x)=0
• \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi
• 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ]
• 3\tan ^3(A)-\tan (A)=0,\:A\in \:[0,\:360]
• 2\cos ^2(x)-\sqrt{3}\cos (x)=0,\:0^{\circ \:}\lt x\lt 360^{\circ \:}
• What is tangent equal to?
• The tangent function (tan), is a trigonometric function that relates the ratio of the length of the side opposite a given angle in a right-angled triangle to the length of the side adjacent to that angle.
• How to solve trigonometric equations step-by-step?
• To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions.
• What is a basic trigonometric equation?
• A basic trigonometric equation has the form sin(x)=a, cos(x)=a, tan(x)=a, cot(x)=a
• How to convert radians to degrees?
• The formula to convert radians to degrees: degrees = radians * 180 / π
• What is cotangent equal to?
• The cotangent function (cot(x)), is the reciprocal of the tangent function.cot(x) = cos(x) / sin(x)

trigonometric-equation-calculator

• Spinning The Unit Circle (Evaluating Trig Functions ) If you’ve ever taken a ferris wheel ride then you know about periodic motion, you go up and down over and over...

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## Solving Trigonometric Equations Part III Lesson

• Learn how to solve trigonometric equations using square roots
• Learn how to solve trigonometric equations using squaring
• Learn how to solve trigonometric equations using identities

## How to Solve Trigonometric Equations Using Square Roots, Squaring, & Identities

Unit circle, solving trigonometric equations by taking square roots, solving trigonometric equations using trigonometric identities, solving trigonometric equations using squaring, skills check:.

Solve each equation for 0 ≤ θ < 2π $$3\text{sec}^2 θ=4$$

Solve each equation for 0 ≤ θ < 2π $$\text{csc}\hspace{.1em}θ + \text{cot}\hspace{.1em}θ + 4=5$$

Solve each equation for 0 ≤ θ < 2π $$-2 + 4\text{cos}\hspace{.1em}θ + \text{sin}^2 θ=3 \text{cos}^2 θ$$

Congrats, Your Score is 100 %

Better Luck Next Time, Your Score is %

Ready for more? Watch the Step by Step Video Lesson   |   Take the Practice Test

## 7.5 Solving Trigonometric Equations

Learning objectives.

In this section, you will:

• Solve linear trigonometric equations in sine and cosine.
• Solve equations involving a single trigonometric function.
• Solve trigonometric equations using a calculator.
• Solve trigonometric equations that are quadratic in form.
• Solve trigonometric equations using fundamental identities.
• Solve trigonometric equations with multiple angles.
• Solve right triangle problems.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

## Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2 π . 2 π . In other words, every 2 π 2 π units, the y- values repeat. If we need to find all possible solutions, then we must add 2 π k , 2 π k , where k k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2 π : 2 π :

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

## Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation cos θ = 1 2 . cos θ = 1 2 .

From the unit circle , we know that

These are the solutions in the interval [ 0 , 2 π ] . [ 0 , 2 π ] . All possible solutions are given by

where k k is an integer.

## Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = 1 2 . sin t = 1 2 .

Solving for all possible values of t means that solutions include angles beyond the period of 2 π . 2 π . From Figure 2 , we can see that the solutions are π 6 π 6 and 5 π 6 . 5 π 6 . But the problem is asking for all possible values that solve the equation. Therefore, the answer is

Given a trigonometric equation, solve using algebra .

• Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
• Substitute the trigonometric expression with a single variable, such as x x or u . u .
• Solve the equation the same way an algebraic equation would be solved.
• Substitute the trigonometric expression back in for the variable in the resulting expressions.
• Solve for the angle.

## Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π . 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π .

Use algebraic techniques to solve the equation.

Solve exactly the following linear equation on the interval [ 0 , 2 π ) : 2 sin x + 1 = 0. [ 0 , 2 π ) : 2 sin x + 1 = 0.

## Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2 ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π , π , not 2 π . 2 π . Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π 2 , π 2 , unless, of course, a problem places its own restrictions on the domain.

## Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π .

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ . sin θ . Then we will find the angles.

## Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: csc θ = − 2 , 0 ≤ θ < 4 π . csc θ = − 2 , 0 ≤ θ < 4 π .

We want all values of θ θ for which csc θ = − 2 csc θ = − 2 over the interval 0 ≤ θ < 4 π . 0 ≤ θ < 4 π .

As sin θ = − 1 2 , sin θ = − 1 2 , notice that all four solutions are in the third and fourth quadrants.

## Solving an Equation Involving Tangent

Solve the equation exactly: tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π . tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π .

Recall that the tangent function has a period of π . π . On the interval [ 0 , π ) , [ 0 , π ) , and at the angle of π 4 , π 4 , the tangent has a value of 1. However, the angle we want is ( θ − π 2 ) . ( θ − π 2 ) . Thus, if tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , then

Over the interval [ 0 , 2 π ) , [ 0 , 2 π ) , we have two solutions:

Find all solutions for tan x = 3 . tan x = 3 .

## Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π . 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π .

We can solve this equation using only algebra. Isolate the expression tan x tan x on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of −1 : θ = 3 π 4 −1 : θ = 3 π 4 and θ = 7 π 4 . θ = 7 π 4 .

## Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

## Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sin θ = 0.8 , sin θ = 0.8 , where θ θ is in radians.

Make sure mode is set to radians. To find θ , θ , use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the sin − 1 sin − 1 function. What is shown on the screen is sin − 1 ( . sin − 1 ( . The calculator is ready for the input within the parentheses. For this problem, we enter sin − 1 ( 0.8 ) , sin − 1 ( 0.8 ) , and press ENTER. Thus, to four decimals places,

The solution is

The angle measurement in degrees is

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ . π − θ .

## Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation sec θ = −4 , sec θ = −4 , giving your answer in radians.

We can begin with some algebra.

Check that the MODE is in radians. Now use the inverse cosine function.

Since π 2 ≈ 1.57 π 2 ≈ 1.57 and π ≈ 3.14 , π ≈ 3.14 , 1.8235 is between these two numbers, thus θ ≈ 1 .8235 θ ≈ 1 .8235 is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 2 .

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is θ ' ≈ π − 1 .8235 ≈ 1 .3181 . θ ' ≈ π − 1 .8235 ≈ 1 .3181 . The other solution in quadrant III is π + 1 .3181 ≈ 4 .4597 . π + 1 .3181 ≈ 4 .4597 .

The solutions are 1.8235 ± 2 π k 1.8235 ± 2 π k and 4.4597 ± 2 π k . 4.4597 ± 2 π k .

Solve cos θ = − 0.2. cos θ = − 0.2.

## Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x x or u . u . If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

## Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π . cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π .

We begin by using substitution and replacing cos θ θ with x . x . It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x . cos θ = x . We have

The equation cannot be factored, so we will use the quadratic formula x = − b ± b 2 − 4 a c 2 a . x = − b ± b 2 − 4 a c 2 a .

Replace x x with cos θ , cos θ , and solve. Thus,

Note that only the + sign is used. This is because we get an error when we solve θ = cos − 1 ( − 3 − 13 2 ) θ = cos − 1 ( − 3 − 13 2 ) on a calculator, since the domain of the inverse cosine function is [ − 1 , 1 ] . [ − 1 , 1 ] . However, there is a second solution:

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

## Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π . 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π .

Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u , sin θ = u , or imagine it, as we factor:

Now set each factor equal to zero.

Next solve for θ : sin θ ≠ 3 2 , θ : sin θ ≠ 3 2 , as the range of the sine function is [ −1 , 1 ] . [ −1 , 1 ] . However, sin θ = 1 , sin θ = 1 , giving the solution π 2 . π 2 .

Make sure to check all solutions on the given domain as some factors have no solution.

Solve sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . [Hint: Make a substitution to express the equation only in terms of cosine.]

## Solving a Trigonometric Equation Using Algebra

Solve exactly:

This problem should appear familiar as it is similar to a quadratic. Let sin θ = x . sin θ = x . The equation becomes 2 x 2 + x = 0. 2 x 2 + x = 0. We begin by factoring:

Set each factor equal to zero.

Then, substitute back into the equation the original expression sin θ sin θ for x . x . Thus,

The solutions within the domain 0 ≤ θ < 2 π 0 ≤ θ < 2 π are 0 , π , 7 π 6 , 11 π 6 . 0 , π , 7 π 6 , 11 π 6 .

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

We can see the solutions on the graph in Figure 3 . On the interval 0 ≤ θ < 2 π , 0 ≤ θ < 2 π , the graph crosses the x- axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in Figure 2 as well.

## Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π .

We can factor using grouping. Solution values of θ θ can be found on the unit circle:

Solve the quadratic equation 2 cos 2 θ + cos θ = 0. 2 cos 2 θ + cos θ = 0.

## Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

## Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2 π . 0 ≤ x < 2 π .

Notice that the left side of the equation is the difference formula for cosine.

From the unit circle in Figure 2 , we see that cos x = 3 2 cos x = 3 2 when x = π 6 , 11 π 6 . x = π 6 , 11 π 6 .

## Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: cos ( 2 θ ) = cos θ . cos ( 2 θ ) = cos θ .

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

So, if cos θ = − 1 2 , cos θ = − 1 2 , then θ = 2 π 3 ± 2 π k θ = 2 π 3 ± 2 π k and θ = 4 π 3 ± 2 π k ; θ = 4 π 3 ± 2 π k ; if cos θ = 1 , cos θ = 1 , then θ = 0 ± 2 π k . θ = 0 ± 2 π k .

## Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π . 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π .

If we rewrite the right side, we can write the equation in terms of cosine:

Our solutions are 2 π 3 , 4 π 3 , π . 2 π 3 , 4 π 3 , π .

## Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin ( 2 x ) sin ( 2 x ) or cos ( 3 x ) . cos ( 3 x ) . When confronted with these equations, recall that y = sin ( 2 x ) y = sin ( 2 x ) is a horizontal compression by a factor of 2 of the function y = sin x . y = sin x . On an interval of 2 π , 2 π , we can graph two periods of y = sin ( 2 x ) , y = sin ( 2 x ) , as opposed to one cycle of y = sin x . y = sin x . This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to sin ( 2 x ) = 0 sin ( 2 x ) = 0 compared to sin x = 0. sin x = 0. This information will help us solve the equation.

## Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos ( 2 x ) = 1 2 cos ( 2 x ) = 1 2 on [ 0 , 2 π ) . [ 0 , 2 π ) .

We can see that this equation is the standard equation with a multiple of an angle. If cos ( α ) = 1 2 , cos ( α ) = 1 2 , we know α α is in quadrants I and IV. While θ = cos − 1 1 2 θ = cos − 1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 cos θ = 1 2 will be in quadrants I and IV.

Therefore, the possible angles are θ = π 3 θ = π 3 and θ = 5 π 3 . θ = 5 π 3 . So, 2 x = π 3 2 x = π 3 or 2 x = 5 π 3 , 2 x = 5 π 3 , which means that x = π 6 x = π 6 or x = 5 π 6 . x = 5 π 6 . Does this make sense? Yes, because cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 . cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 .

In quadrant I, 2 x = π 3 , 2 x = π 3 , so x = π 6 x = π 6 as noted. Let us revolve around the circle again:

so x = 7 π 6 . x = 7 π 6 .

One more rotation yields

x = 13 π 6 > 2 π , x = 13 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

In quadrant IV, 2 x = 5 π 3 , 2 x = 5 π 3 , so x = 5 π 6 x = 5 π 6 as noted. Let us revolve around the circle again:

so x = 11 π 6 . x = 11 π 6 .

x = 17 π 6 > 2 π , x = 17 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

Our solutions are π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . Note that whenever we solve a problem in the form of sin ( n x ) = c , sin ( n x ) = c , we must go around the unit circle n n times.

## Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem . We begin with the familiar Pythagorean Theorem, a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , and model an equation to fit a situation.

## Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4 .

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

The angle of elevation is θ , θ , formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

The angle of elevation is approximately 71.7 ∘ , 71.7 ∘ , and the length of the cable is 73.2 meters.

## Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be 4 a feet. See Figure 5 .

The side adjacent to θ θ is a and the hypotenuse is 4 a . 4 a . Thus,

The elevation of the ladder forms an angle of 75.5 ∘ 75.5 ∘ with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

Thus, the ladder touches the wall at 15 a 15 a feet from the ground.

Access these online resources for additional instruction and practice with solving trigonometric equations.

• Solving Trigonometric Equations I
• Solving Trigonometric Equations II
• Solving Trigonometric Equations III
• Solving Trigonometric Equations IV
• Solving Trigonometric Equations V
• Solving Trigonometric Equations VI

## 7.5 Section Exercises

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?

For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2 π . 0 ≤ θ < 2 π .

2 sin θ = − 2 2 sin θ = − 2

2 sin θ = 3 2 sin θ = 3

2 cos θ = 1 2 cos θ = 1

2 cos θ = − 2 2 cos θ = − 2

tan θ = −1 tan θ = −1

tan x = 1 tan x = 1

cot x + 1 = 0 cot x + 1 = 0

4 sin 2 x − 2 = 0 4 sin 2 x − 2 = 0

csc 2 x − 4 = 0 csc 2 x − 4 = 0

For the following exercises, solve exactly on [ 0 , 2 π ) . [ 0 , 2 π ) .

2 cos θ = 2 2 cos θ = 2

2 cos θ = −1 2 cos θ = −1

2 sin θ = −1 2 sin θ = −1

2 sin θ = − 3 2 sin θ = − 3

2 sin ( 3 θ ) = 1 2 sin ( 3 θ ) = 1

2 sin ( 2 θ ) = 3 2 sin ( 2 θ ) = 3

2 cos ( 3 θ ) = − 2 2 cos ( 3 θ ) = − 2

cos ( 2 θ ) = − 3 2 cos ( 2 θ ) = − 3 2

2 sin ( π θ ) = 1 2 sin ( π θ ) = 1

2 cos ( π 5 θ ) = 3 2 cos ( π 5 θ ) = 3

For the following exercises, find all exact solutions on [ 0 , 2 π ) . [ 0 , 2 π ) .

sec ( x ) sin ( x ) − 2 sin ( x ) = 0 sec ( x ) sin ( x ) − 2 sin ( x ) = 0

tan ( x ) − 2 sin ( x ) tan ( x ) = 0 tan ( x ) − 2 sin ( x ) tan ( x ) = 0

2 cos 2 t + cos ( t ) = 1 2 cos 2 t + cos ( t ) = 1

2 tan 2 ( t ) = 3 sec ( t ) 2 tan 2 ( t ) = 3 sec ( t )

2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0 2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0

cos 2 θ = 1 2 cos 2 θ = 1 2

sec 2 x = 1 sec 2 x = 1

tan 2 ( x ) = − 1 + 2 tan ( − x ) tan 2 ( x ) = − 1 + 2 tan ( − x )

8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0 8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0

tan 5 ( x ) = tan ( x ) tan 5 ( x ) = tan ( x )

For the following exercises, solve with the methods shown in this section exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9 sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9

sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1 sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1

cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1 cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1

6 sin ( 2 t ) + 9 sin t = 0 6 sin ( 2 t ) + 9 sin t = 0

9 cos ( 2 θ ) = 9 cos 2 θ − 4 9 cos ( 2 θ ) = 9 cos 2 θ − 4

sin ( 2 t ) = cos t sin ( 2 t ) = cos t

cos ( 2 t ) = sin t cos ( 2 t ) = sin t

cos ( 6 x ) − cos ( 3 x ) = 0 cos ( 6 x ) − cos ( 3 x ) = 0

For the following exercises, solve exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Use the quadratic formula if the equations do not factor.

tan 2 x − 3 tan x = 0 tan 2 x − 3 tan x = 0

sin 2 x + sin x − 2 = 0 sin 2 x + sin x − 2 = 0

sin 2 x − 2 sin x − 4 = 0 sin 2 x − 2 sin x − 4 = 0

5 cos 2 x + 3 cos x − 1 = 0 5 cos 2 x + 3 cos x − 1 = 0

3 cos 2 x − 2 cos x − 2 = 0 3 cos 2 x − 2 cos x − 2 = 0

5 sin 2 x + 2 sin x − 1 = 0 5 sin 2 x + 2 sin x − 1 = 0

tan 2 x + 5 tan x − 1 = 0 tan 2 x + 5 tan x − 1 = 0

cot 2 x = − cot x cot 2 x = − cot x

− tan 2 x − tan x − 2 = 0 − tan 2 x − tan x − 2 = 0

For the following exercises, find exact solutions on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Look for opportunities to use trigonometric identities.

sin 2 x − cos 2 x − sin x = 0 sin 2 x − cos 2 x − sin x = 0

sin 2 x + cos 2 x = 0 sin 2 x + cos 2 x = 0

sin ( 2 x ) − sin x = 0 sin ( 2 x ) − sin x = 0

cos ( 2 x ) − cos x = 0 cos ( 2 x ) − cos x = 0

2 tan x 2 − sec 2 x − sin 2 x = cos 2 x 2 tan x 2 − sec 2 x − sin 2 x = cos 2 x

1 − cos ( 2 x ) = 1 + cos ( 2 x ) 1 − cos ( 2 x ) = 1 + cos ( 2 x )

sec 2 x = 7 sec 2 x = 7

10 sin x cos x = 6 cos x 10 sin x cos x = 6 cos x

−3 sin t = 15 cos t sin t −3 sin t = 15 cos t sin t

4 cos 2 x − 4 = 15 cos x 4 cos 2 x − 4 = 15 cos x

8 sin 2 x + 6 sin x + 1 = 0 8 sin 2 x + 6 sin x + 1 = 0

8 cos 2 θ = 3 − 2 cos θ 8 cos 2 θ = 3 − 2 cos θ

6 cos 2 x + 7 sin x − 8 = 0 6 cos 2 x + 7 sin x − 8 = 0

12 sin 2 t + cos t − 6 = 0 12 sin 2 t + cos t − 6 = 0

tan x = 3 sin x tan x = 3 sin x

cos 3 t = cos t cos 3 t = cos t

For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros.

6 sin 2 x − 5 sin x + 1 = 0 6 sin 2 x − 5 sin x + 1 = 0

8 cos 2 x − 2 cos x − 1 = 0 8 cos 2 x − 2 cos x − 1 = 0

100 tan 2 x + 20 tan x − 3 = 0 100 tan 2 x + 20 tan x − 3 = 0

2 cos 2 x − cos x + 15 = 0 2 cos 2 x − cos x + 15 = 0

20 sin 2 x − 27 sin x + 7 = 0 20 sin 2 x − 27 sin x + 7 = 0

2 tan 2 x + 7 tan x + 6 = 0 2 tan 2 x + 7 tan x + 6 = 0

130 tan 2 x + 69 tan x − 130 = 0 130 tan 2 x + 69 tan x − 130 = 0

For the following exercises, use a calculator to find all solutions to four decimal places.

sin x = 0.27 sin x = 0.27

sin x = −0.55 sin x = −0.55

tan x = −0.34 tan x = −0.34

cos x = 0.71 cos x = 0.71

For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Round to four decimal places.

tan 2 x + 3 tan x − 3 = 0 tan 2 x + 3 tan x − 3 = 0

6 tan 2 x + 13 tan x = −6 6 tan 2 x + 13 tan x = −6

tan 2 x − sec x = 1 tan 2 x − sec x = 1

sin 2 x − 2 cos 2 x = 0 sin 2 x − 2 cos 2 x = 0

2 tan 2 x + 9 tan x − 6 = 0 2 tan 2 x + 9 tan x − 6 = 0

4 sin 2 x + sin ( 2 x ) sec x − 3 = 0 4 sin 2 x + sin ( 2 x ) sec x − 3 = 0

For the following exercises, find all solutions exactly to the equations on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

csc 2 x − 3 csc x − 4 = 0 csc 2 x − 3 csc x − 4 = 0

sin 2 x − cos 2 x − 1 = 0 sin 2 x − cos 2 x − 1 = 0

sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0 sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0

3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0 3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0

sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1 sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1

tan 2 x − 1 − sec 3 x cos x = 0 tan 2 x − 1 − sec 3 x cos x = 0

sin ( 2 x ) sec 2 x = 0 sin ( 2 x ) sec 2 x = 0

sin ( 2 x ) 2 csc 2 x = 0 sin ( 2 x ) 2 csc 2 x = 0

2 cos 2 x − sin 2 x − cos x − 5 = 0 2 cos 2 x − sin 2 x − cos x − 5 = 0

1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4 1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4

## Real-World Applications

An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly?

If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground?

If a loading ramp is placed next to a truck, at a height of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground?

A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal?

An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.)

A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building?

A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him?

A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun?

A 90-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun?

A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light?

A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light?

For the following exercises, find a solution to the word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree.

A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall?

A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall?

A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping?

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• Book title: Precalculus 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
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• Section URL: https://openstax.org/books/precalculus-2e/pages/7-5-solving-trigonometric-equations

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## How to Solve Trigonometric Equations

Summary: A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x  = cos  x , can be solved only numerically, through successive approximations. But a great many can be solved analytically — in “closed form”, an exact solution in symbols — and this page shows you how to do it in five steps.

## Step 1. Get one function of one angle.

Step 2. solve for the value(s) of a trig function., step 3. solve for the angle., step 4. solve for the variable., step 5. apply any restrictions., more examples.

If a trig equation can be solved analytically, these steps will do it:

On this page I’ll walk you through solving several trig equations using these steps, showing you every detail. Once you know and understand the steps, you’ll be able to work some more examples more quickly.

Trig equations, like any equations, are really about numbers, not angles. You are looking for all possible numbers that could be substituted for the variable in the equation to make it true. But it simplifies things to think about the angles first and worry about the variables later.

Example A :

cos(4 A ) − sin(2 A ) = 0

Here the “angles”, the arguments to the trig functions, are 4 A and 2 A . True, you want to solve for A ultimately. But if you can solve for the angle 4 A or 2 A , it is then quite easy to solve for the variable .

As you see, that equation involves two functions (sine and cosine) of two angles (4 A and 2 A ). You need to get it in terms of one function of one angle. Note well: a function of one angle, not necessarily a function of just the variable A .

This is where it is essential to have a nodding acquaintance with all the trig identities . If you do, you’ll remember that cos(2 u ) can be expressed in terms of sin( u ). Specifically, cos(2 u ) = 1 − 2sin²( u ).

How does that help? Well, 4 A is 2×2 A , isn’t it?

cos(2 u ) = 1 − 2sin²( u )

cos(2×2 A ) = 1 − 2sin²(2 A )

cos(4 A ) = 1 − 2sin²(2 A )

That transforms the original equation to

1 − 2sin²(2 A ) − sin(2 A ) = 0

which can be rewritten in standard form as

2sin²(2 A ) + sin(2 A ) − 1 = 0

Now you have the equation in terms of only one function (sine) and only one angle (2 A ).

Now that the equation involves only a single function of a single angle, your next task is to solve for that function value.

Example A continues. Recapping what was done so far,

cos(4 A ) − sin(2 A ) = 0  ⇒

You want to solve for sin(2 A ). You should recognize that the equation is really a quadratic,

2 y ² + y − 1 = 0, where y = sin(2 A )

It can be factored in a straightforward way, as ( y +1)(2 y −1), which gives

(sin(2 A ) + 1) (2sin(2 A ) − 1) = 0

From algebra you know that if a product is 0 then you solve by setting each factor to 0:

sin(2 A ) + 1 = 0       or       2sin(2 A ) − 1 = 0

sin(2 A ) = −1       or       sin(2 A ) = 1/2

Example B :

3tan²( B /2) − 1 = 0

tan²( B /2) = 1/3

and then take square root of both sides:

tan( B /2) = ±√ 1/3  = ±√ 3 /3

It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.

After solving for a function value, now you solve for the angle. If it’s a multiple of π/6 (30°) or π/4 (45°), you can easily solve it exactly. If it’s a half-multiple of those angles, you may be able to use the half-angle formulas to write an exact solution. Otherwise, you’ll need to write the solution as an arcfunction.

Trig equations have one important difference from other types of equations. Trig functions are periodic , meaning that they repeat their values over and over. Therefore a trig equation has an infinite number of solutions if it has any.

Think about an equation like sin  u  = 1. π/2 is a solution, but the sine function repeats its values every 2π. Therefore π/2±2π, π/2±4π, and so on are equally good solutions. To show this, write the solution as u  = π/2 + 2π n , where n is understood to be any integer, positive, negative, or zero. (The tangent and cotangent functions repeat all their values every π radians, so the solution to tan  v  = 1 is v  = π/4 + π n , not +2π n .)

2sin²(2 A ) + sin(2 A ) − 1 = 0 ⇒

The sine of 3π/2 is −1, so the first possibility reduces to 2 A  = 3π/2. But remember that the sine function is periodic, so write

sin(2 A ) = −1 ⇒ 2 A  = 3π/2 + 2π n .

For the second possibility, sin(2 A ) = 1/2, there are two solutions, because sin(π/6) and sin(5π/6) both equal 1/2, and again we add 2π n to the angle to account for all solutions:

sin(2 A ) = 1/2 ⇒ 2 A = π/6 + 2π n or 5π/6 + 2π n

Combining these, here are the three solutions for the original equation:

2 A = 3π/2 + 2π n    or    π/6 + 2π n     or    5π/6 + 2π n

Example B continues. Recapping what was done so far,

3tan²( B /2) − 1 = 0 ⇒

tan( B /2) = ±√ 3 /3

What angle has a tangent value of √ 3 /3? the angle π/6. And where does the tangent have a value of −√ 3 /3? at the angle 5π/6. This gives the solutions

B /2 = π/6 + π n    or     5π/6 + π n

Remember that the tangent and cotangent have period π and not 2π.

Example C :

Of course, you don’t always luck out with nice angles. Take a look at this equation:

sec(3 C ) = 2.5

1/sec(3 C ) = 1/2.5

cos(3 C ) = 0.4

For what angles is that true? We write arccos(0.4) to mean the angle in quadrant I that has a cosine equal to 0.4. (Some books write cos −1 (0.4) instead of arccos(0.4). I prefer the arccos notation because the superscript −1 makes many students think of 1/cos(0.4), which has a different meaning entirely.)

So initially we would write 3 C  = arccos(0.4) + 2π n . But that’s not the whole story: any angle in quadrant I has a reflection in quadrant IV with the same cosine value, so we need to account for both angles:

3 C = arccos(0.4) + 2π n   or   2π−arccos(0.4) + 2π n

Note that the base angle is always nonnegative and less than 2π: 2π−arccos(0.4) + 2π n , not simply −arccos(0.4) + 2π n . This is necessary to make step 5 come out right.

Now it’s time to abandon angular thinking and go for the variable. As you will see, it is very important to do this step after step 3 , not before. 2π n or π n must be added to the angle, not the variable, to reflect the period of the trig function.

sin(2 A ) = −1   or   sin(2 A ) = 1/2 ⇒

2 A = 3π/2 + 2π n   or   π/6 + 2π n    or   5π/6 + 2π n

Now divide both sides by 2 to solve for the variable, A :

A = 3π/4 + π n   or   π/12 + π n   or   5π/12 + π n

Be sure to divide the entire equation, so that the 2π n becomes π n .

Why is the order of steps so important? The 2πn came in because the sine function has a period of 2π: if you take an angle and add 2π to it, it looks like the same angle and all six of its function values are unchanged. But now we’re no longer dealing with the angle 2 A , we’re dealing with the variable A . In this equation, we say that adding π to any solution for A will give another solution for A .

For instance, set n =1 and obtain solutions A  = 7π/4, 13π/12, or 17π/12. True, sin(7π/4) doesn’t equal −1. But the equation was sin(2 A ) = −1, not sin( A ) = −1. If you substitute A  = 7π/4 in sin(2 A ), you get sin(2·7π/4) = sin(7π/2) which does equal −1. Always pay attention to whether you’re dealing with the angle or the variable.

tan( B /2) = ±√ 3 /3 ⇒

B /2 = π/6 + π n   or   5π/6 + π n

Multiplying both sides by 2 gives

B = π/3 + 2π n   or   5π/3 + 2π n

Once again, the angle was B /2 and had a period of π; the variable is B and has a period of 2π.

Example C continues. Recapping what was done so far,

sec(3 C ) = 2.5 ⇒

Divide both sides by 3:

C = (1/3)arccos(0.4) + (2π/3) n   or   (2π/3)−(1/3)arccos(0.4) + (2π/3) n

It’s a matter of taste whether to combine terms in that second solution:

C = (1/3)arccos(0.4) + (2π/3) n    or    −(1/3)arccos(0.4) + (2 n +1)π/3

Does the problem specify a solution interval for the variable? Sometimes this is done in interval notation, like [0;2π); other times it’s done as an inequality, 0 <=  x  < 2π. If no restriction is given, you should give all real solutions as shown in step 4 .

But if solutions are restricted to a particular interval, you have a bit more work to do after solving for the variable .

2 A = 3π/2 + 2π n   or   π/6 + 2π n    or   5π/6 + 2π n ⇒

Now suppose that only solutions on the interval [0;2π) were wanted.

The general solutions have a period of π (from +π n ), and therefore there will be two cycles between 0 and 2π:

The solutions for n  = 2 are all larger than the 2π boundary of the interval. There are six solutions to the equation for A within the interval [0;2π). In order, they are π/12, 5π/12, 3π/4, 13π/12, 17π/12, and 7π/4.

B /2 = π/6 + π n   or   5π/6 + π n ⇒

Suppose that the problem specified solutions between 0 and π/2. As you can see, even with n  = 0 there is just one solution within the limits [0;π/2). Answer: π/3 only.

3 C = arccos(0.4) + 2π n   or   2π−arccos(0.4) + 2π n ⇒

Suppose again that limits of C in [0;2π) are stated. The period of the variable is 2π/3, so there are three cycles ( n =0,1,2) between 0 and 2π. Here are the values, with decimal approximations:

Since 2π is about 6.28, there are six solutions within the stated limits. C  = about 0.3864, 0.6608, 2.4808, 2.7552, 4.5752, and 4.8496.

Example D :

Solve ½sin(2 D ) − ½ = 0 for D in [0;2π).

Solution: There is only one function (sin) of one angle (2 D ), so step 1 is complete. Don’t “simplify” ½sin(2 D ) to sin( D ): that is not a valid operation . And don’t convert sin(2 D ) to 2 sin( D ) cos( D ); while that is mathematically valid, it makes the equation more complicated, not simpler.

Step 2 : solve for the function.

½sin(2 D ) = ½

sin(2 D ) = 1

Step 3 : find the angle.

2 D = π/2 + 2π n

Step 4 : solve for the variable.

D = π/4 + π n

Step 5 : apply any restrictions. The period of the variable is π, which means there will be two cycles in the interval [0;2π).

Example E :

Find all solutions for sin²( E /2) − cos²( E /2) = 1.

Solution: In step 1 , you use identities to get the equation in term of one function of one angle. As often happens in trig, you have a choice of which identity you want to use. You could replace cos²( E /2) with 1−sin²( E /2), or you could use the double-angle formula cos(2 u ) = cos²( u ) − sin²( u ) with u  =  E /2. That second approach leads to a simpler equation:

sin²( E /2) − cos²( E /2) = 1

cos²( E /2) − sin²( E /2) = −1

cos(2 E /2) = −1

cos( E ) = −1

Step 3 : write down the angle:

E = π + 2π n = (2 n +1)π

Step 4 is already done, since the angle is the variable.

Step 5 is easily done: the problem specifically asks for all values of E .

Answer: E  =(2 n +1)π for any integer n .

Showed an intermediate step in a “straightforward” solution.

Add the possibility of angles being multiples of half of a special angle .

Converted page from HTML 4.01 to HTML5, and aitalicized the variable names.

• 17 Aug 2015 : Moved from OakRoadSystems.com to BrownMath.com .
• (intervening changes suppressed)
• 20 Apr 2002 : First publication.

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## How to Easily Solve Trigonometric Equations

Last Updated: February 1, 2023 Fact Checked

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 17 people, some anonymous, worked to edit and improve it over time. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 258,444 times. Learn more...

Did you get homework from your teacher that was about solving Trigonometric equations? Did you maybe not pay full attention in class during the lesson on Trigonometric questions? Do you even know what "Trigonometric" means? If you answered yes to these questions, then you don't need to worry because this wikiHow will teach you how to solve Trigonometric equations.

• To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations.

• There are 4 types of basic trig equations:
• sin x = a ; cos x = a
• tan x = a ; cot x = a
• Solving basic trig equations proceeds by studying the various positions of the arc x on the trig circle, and by using trig conversion table (or calculator ). To fully know how to solve these basic trig equations, and similar, see book titled :"Trigonometry: Solving trig equations and inequalities" (Amazon E-book 2010).
• Example 1. Solve sin x = 0.866. The conversion table (or calculator) gives the answer: x = Pi/3. The trig circle gives another arc (2Pi/3) that has the same sin value (0.866). The trig circle also gives an infinity of answers that are called extended answers.
• x1 = Pi/3 + 2k.Pi, and x2 = 2Pi/3. (Answers within period (0, 2Pi))
• x1 = Pi/3 + 2k Pi, and x2 = 2Pi/3 + 2k Pi. (Extended answers).
• Example 2. Solve: cos x = -1/2. Calculators give x = 2 Pi/3. The trig circle gives another x = -2Pi/3.
• x1 = 2Pi/3 + 2k.Pi, and x2 = - 2Pi/3. (Answers within period (0, 2Pi))
• x1 = 2Pi/3 + 2k Pi, and x2 = -2Pi/3 + 2k.Pi. (Extended answers)
• Example 3. Solve: tan (x - Pi/4) = 0.
• x = Pi/4 ; (Answer)
• x = Pi/4 + k Pi; ( Extended answer)
• Example 4. Solve cot 2x = 1.732. Calculators and the trig circle give
• x = Pi/12 ; (Answer)
• x = Pi/12 + k Pi ; (Extended answers)

• To transform a given trig equation into basic trig ones, use common algebraic transformations ( factoring , common factor , polynomial identities...), definitions and properties of trig functions, and trig identities. There are about 31, among them the last 14 trig identities, from 19 to 31, are called Transformation Identities, since they are used in the transformation of trig equations. [4] X Research source See book mentioned above.
• Example 5: The trig equation: sin x + sin 2x + sin 3x = 0 can be transformed, using trig identities, into a product of basic trig equations: 4cos x*sin (3x/2)*cos (x/2) = 0. The basic trig equations to be solved are: cos x = 0 ; sin (3x/2) = 0 ; and cos (x/2) = 0.

• Before learning solving trig equations, you must know how to quickly find the arcs whose trig functions are known. Conversion values of arcs (or angles) are given by trig tables or calculators. [6] X Research source
• Example: After solving, get cos x = 0.732. Calculators give the solution arc x = 42.95 degree. The trig unit circle will give other solution arcs that have the same cos value.

• You can graph to illustrate the solution arcs on the trig unit circle. The terminal points of these solution arcs constitute regular polygons on the trig circle. For examples:
• The terminal points of the solution arcs x = Pi/3 + k.Pi/2 constitute a square on the trig unit circle.
• The solution arcs x = Pi/4 + k.Pi/3 are represented by the vertexes of a regular hexagon on the trig unit circle.

• A. Approach 1.
• Transform the given trig equation into a product in the form: f(x).g(x) = 0 or f(x).g(x).h(x) = 0, in which f(x), g(x) and h(x) are basic trig equations.
• Example 6. Solve: 2cos x + sin 2x = 0. (0 < x < 2Pi)
• Solution. Replace in the equation sin 2x by using the identity: sin 2x = 2*sin x*cos x.
• cos x + 2*sin x*cos x = 2cos x*( sin x + 1) = 0. Next, solve the 2 basic trig functions: cos x = 0, and (sin x + 1) = 0.
• Example 7. Solve: cos x + cos 2x + cos 3x = 0. (0 < x < 2Pi)
• Solution: Transform it to a product, using trig identities: cos 2x(2cos x + 1 ) = 0. Next, solve the 2 basic trig equations: cos 2x = 0, and (2cos x + 1) = 0.
• Example 8. Solve: sin x - sin 3x = cos 2x. (0 < x < 2Pi)
• B. Approach 2.
• Transform the given trig equation into a trig equation having only one unique trig function as variable. There are a few tips on how to select the appropriate variable. The common variables to select are: sin x = t; cos x = t; cos 2x = t, tan x = t and tan (x/2) = t.
• Example 9. Solve: 3sin^2 x - 2cos^2 x = 4sin x + 7 (0 < x < 2Pi).
• Solution. Replace in the equation (cos^2 x) by (1 - sin^2 x), then simplify the equation:
• 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Call sin x = t. The equation becomes: 5t^2 - 4t - 9 = 0. This is a quadratic equation that has 2 real roots: t1 = -1 and t2 = 9/5. The second t2 is rejected since > 1. Next, solve: t = sin = -1 --> x = 3Pi/2.
• Example 10. Solve: tan x + 2 tan^2 x = cot x + 2.
• Solution. Call tan x = t. Transform the given equation into an equation with t as variable: (2t + 1)(t^2 - 1) = 0. Solve for t from this product, then solve the basic trig equation tan x = t for x.

• There are a few special types of trig equations that require some specific transformations. Examples:
• a*sin x+ b*cos x = c ; a(sin x + cos x) + b*cos x*sin x = c ;
• a*sin^2 x + b*sin x*cos x + c*cos^2 x = 0

• The function f(x) = sin x has 2Pi as period.
• The function f(x) = tan x has Pi as period.
• The function f(x) = sin 2x has Pi as period.
• The function f(x) = cos (x/2) has 4Pi as period.
• If the period is specified in the problem/test, you have to only find the solution arc(s) x within this period.
• NOTE: Solving trig equation is a tricky work that often leads to errors and mistakes. Therefore, answers should be carefully checked. After solving, you can check the answers by using a graphing calculator to directly graph the given trig equation R(x) = 0. The answers (real roots) will be given in decimals. For example, Pi is given by the value 3.14
• ↑ https://www.purplemath.com/modules/solvtrig.htm
• ↑ https://www.mathopenref.com/arcsin.html
• ↑ https://courses.lumenlearning.com/precalculus/chapter/unit-circle-sine-and-cosine-functions/
• ↑ https://mathbitsnotebook.com/Algebra2/TrigConcepts/TCEquationsMore.html
• ↑ https://www.analyzemath.com/trigonometry/properties.html

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## Trigonometric Equations

• Sahithi venkatesan

To solve a trigonometric equation, we need the following preliminary knowledge:

If $$\sin \theta = \sin \alpha$$, then $$\theta=n\pi+(-1)^{n}\alpha$$. Thus, if $$n$$ is odd, $$\theta=(2m+1)\pi-\alpha,$$ and if $$n$$ is even, $$\theta=2m\pi+\alpha$$.

If $$\cos \theta = \cos \alpha$$, then $$\theta = 2n\pi\pm\alpha$$.

If $$\tan \theta = \tan\alpha$$, then $$\theta=n\pi+\alpha$$.

These hold true for integers $$n,m$$.

Now on to solving equations. The general method of solving an equation is to convert it into the form of one ratio only. Then, using these results, we can obtain solutions.

Solving basic equations can be taken care of with the trigonometric R method .

Consider the following example:

Solve the following equation: $\cos x - \sqrt{3}\sin x = 2.$ Using the R method, this equation can be converted into $\cos\left(x+\frac{\pi}{3}\right)=1.$ Now, we know that $$\cos0^ \circ=1$$. Hence we rewrite the equation as $\cos\left(x+\frac{\pi}{3}\right)=\cos0^\circ.$ Using the above results, we have \begin{align} x+\frac{\pi}{3} &= 2n\pi\pm0\\ x&=2n\pi-\frac{\pi}{3}. \ _\square \end{align} Note: Sometimes the question may also provide a range for $$x$$. In this case, we take various integral values for $$x$$ and find the solutions.

But as the equations get harder, a variety of techniques come handy: the double- and triple-angle formulas, the sum-to-product formulas, etc. Sometimes the equation is converted into a form where the R method can be used, or sometimes we have quadratic equations in the field.

Let's take another example.

Find the general solution of the equation $$\cos2x-2\tan x+2=0.$$ Note that $$\cos 2x = \frac{1-\tan^{2} x}{1+\tan^{2} x}.$$ Substituting this, the equation has only one variable: $$\tan x$$. We evaluate and then factorize the expression to obtain $(\tan x - 1)\left(2\tan^{2} x + \tan x + 3\right)=0.$ As the equation $$2\tan^{2} x + \tan x + 3 = 0$$ has imaginary roots, we do not consider it. Hence, we have \begin{align} \tan x - 1 &= 0\\ \tan x &= 1\\ x&=\frac{\pi}{4}+n\pi. \ _\square \end{align}

## Specific Solutions - Basic

Specific solutions - intermediate, general solutions - basic, general solutions - intermediate, factoring - basic, factoring - intermediate, problem solving - basic, problem solving - intermediate.

What are the solutions of the equation $\sin 2x = \frac{\sqrt{3}}{2}$ in the $$x$$-interval $$[0, 2\pi] ?$$ Since $$0 \leq 2x \leq 4\pi ,$$ we have \begin{align} 2x&= \frac{\pi}{3}, \frac{2}{3} \pi, \frac{7}{3} \pi, \frac{8}{3} \pi\\\\ \Rightarrow x&= \frac{\pi}{6}, \frac{\pi}{3}, \frac{7}{6} \pi, \frac{4}{3} \pi. \ _\square \end{align}
What are the solutions of $\sin \left(x-\frac{\pi}{3} \right) = -\frac{1}{2}$ in the $$x$$-interval $$[0, 2\pi] ?$$ Since $$-\frac{\pi}{3} \leq x-\frac{\pi}{3} \leq \frac{5}{3} \pi ,$$ we have \begin{align} x-\frac{\pi}{3} &=-\frac{\pi}{6}, \frac{7}{6} \pi\\\\ \Rightarrow x &= \frac{\pi}{6}, \frac{3}{2} \pi. \ _\square \end{align}
What is the general solution of $\tan x = 1 ?$ In the first period $$0 \leq x < \pi,$$ the solution for this equation is $$x= \frac{\pi}{4}.$$ Since the period of $$\tan x$$ is $$\pi,$$ the general solution of the given equation is $x = n \pi + \frac{\pi}{4} . \ _\square$
What is the general solution of $\cos 2x +3\cos x -1=0 ?$ Since $$\cos 2x = 2\cos^2 x -1,$$ we have \begin{align} 2 \cos^2 x -1 + 3\cos x -1 &= 0 \\ (2\cos x -1)(\cos x +2) &= 0\\ \cos x &= \frac{1}{2}. \qquad (\text{since }\lvert \cos x \rvert \leq 1) \end{align} Since the period of $$\cos x$$ is $$2\pi,$$ the general solution for the given equation is $x = 2n \pi \pm \frac{\pi}{3}. \ _\square$
What is the general solution of $\tan 2x - 3\tan x = 0 ?$ Since $$\tan 2x = \frac{2\tan x}{1-\tan^2 x} ,$$ we have \begin{align} \tan 2x - 3\tan x &= 0 \\ \frac{2\tan x}{1-\tan^2 x} -3\tan x &= 0 \\ 2\tan x -3\tan x \left(1-\tan^2 x\right) &= 0 \\ \tan x\left(3\tan^2 x -1\right) &= 0 \\ \Rightarrow \tan x &= 0, \pm \frac{1}{\sqrt{3}}. \end{align} Since the period of $$\tan x$$ is $$\pi,$$ the general solution for the given equation is $x= n \pi, \text{ or } x=n \pi \pm \frac{\pi}{6}. \ _\square$
What are the solutions of the following equation for $$0 \leq x \leq 2\pi:$$ $2\cos^2 x - \sin x -1=0 ?$ Since $$\sin^2 + \cos^2 = 1 ,$$ we rewrite the given equation to obtain \begin{align} 2 \cos^2 x-\sin x -1 &= 0 \\ \Rightarrow 2(1-\sin^2 x)-\sin x -1 &= 0 \\ \left(2\sin x - 1\right)(\sin x + 1) &=0 \\ \sin x &= \frac{1}{2}, -1. \end{align} Since $$0 \leq x \leq 2\pi,$$ for $$\sin x = \frac{1}{2}$$ we have $x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (1)$ For $$\sin x = -1,$$ we have $x = \frac{3}{2} \pi. \qquad (2)$ Thus, from $$(1)$$ and $$(2)$$ we have $x = \frac{\pi}{6}, \frac{5}{6} \pi, \frac{3}{2} \pi. \ _\square$
What are the solutions of the following equation for $$0 \leq x \leq 2\pi:$$ $\sin^ 4 x + \sin^2 x -1 = \cos^4 x + \cos^2 x ?$ Using $$\sin^2 x + \cos^2 x = 1,$$ we have \begin{align} \sin^ 4 x + \sin^2 x -1 &= \cos^4 x + \cos^2 x \\ \sin^4 x - \cos^4 x + \sin^2 x - \cos^2 x -1 &= 0 \\ \left(\sin^2 x + \cos^2 x\right)\left(\sin^2 x - \cos^2 x\right) + \sin^2x - \cos^2 x -1 &= 0 \\ 2\sin^2 x - 2\cos^2 x - 1 &= 0 \\ 2\sin^2 x - 2\left(1-\sin^2 x\right) -1 &= 0 \\ \Rightarrow \sin^2 x &= \frac{3}{4} \\ \sin x &= \pm \frac{\sqrt{3}}{2}. \end{align} Since $$0 \leq x \leq 2\pi,$$ for $$\sin x = \frac{\sqrt{3}}{2}$$ we have $x = \frac{\pi}{3}, \frac{2}{3} \pi . \qquad (1)$ For $$\sin x = -\frac{\sqrt{3}}{2} ,$$ we have $x= \frac{4}{3} \pi, \frac{5}{3} \pi. \qquad (2)$ Thus, from $$(1)$$ and $$(2)$$ the solutions are $x = \frac{\pi}{3}, \frac{2}{3} \pi, \frac{4}{3} \pi, \frac{5}{3} \pi. \ _\square$
What are the solutions of the following equation for $$0 \leq x \leq 2\pi:$$ $\cos^2 x - \sin^2 2x = 0?$ We have \begin{align} \cos^2 x - \sin^2 2x &= 0 \\ \cos^2 x - 4\sin^2 x \cos^2 x &= 0 \qquad ( \text{ since } \sin 2x = 2\sin x \cos x ) \\ \cos^2x(1-4\sin^2 x) &= 0 \\ \Rightarrow \cos x &= 0, ~\sin x = \frac{1}{2}, - \frac{1}{2}. \end{align} Since $$0 \leq x \leq 2\pi,$$ for $$\cos x =0$$ we have $x = \frac{\pi}{2}, \frac{3}{2} \pi. \qquad (1)$ For $$\sin x = \frac{1}{2},$$ we have $x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (2)$ For $$\sin x = -\frac{1}{2},$$ we have $x= \frac{7}{6} \pi, \frac{11}{6} \pi. \qquad (3)$ Thus, from $$(1), (2)$$ and $$(3)$$ the solutions are $x= \frac{\pi}{2}, \frac{3}{2} \pi, \frac{\pi}{6}, \frac{5}{6} \pi, \frac{7}{6} \pi, \frac{11}{6} \pi. \ _\square$
What are the values of $$\alpha$$ in the interval $$[0, 2\pi]$$ such that the quadratic equation $$x^2 + 2x + 2\cos \alpha=0$$ has repeated roots? For the quadratic equation $$x^2 + 2x +2\cos \alpha = 0$$ to have repeated roots, its discriminant $$D$$ must be 0. Thus, \begin{align} \frac{D}{4} = 1-2\cos \alpha &= 0 \\ \cos \alpha &= \frac{1}{2} \\ \Rightarrow \alpha &= \frac{\pi}{3}, \frac{5}{3} \pi. \ _\square \end{align}

$\sin\theta+\cos\theta=\sqrt{2}\sin(90^{\circ}-\theta), \quad \cot\theta = \, ?$

If a root of the equation $1-6\sin x = \sqrt{5-12\sin x}$ is $$\alpha,$$ what is the value of $$\sin \alpha \cos 2\alpha?$$ Squaring both sides, we have \begin{align} (1-6\sin x)^2 &= \left(\sqrt{5-12\sin x}\right)^2 \\ 1-12\sin x + 36\sin^2 x &= 5 - 12 \sin x \\ \sin^2 x &= \frac{1}{9} \\ \Rightarrow \sin x &= \pm \frac{1}{3}. \end{align} Since only $$\sin x = - \frac{1}{3}$$ satisfies $$1-6\sin x = \sqrt{5-12\sin x} ,$$ we have $$\sin \alpha= -\frac{1}{3}.$$ Thus, $$\sin \alpha \cos 2\alpha$$ is \begin{align} \sin \alpha \cos 2\alpha &= \sin \alpha\left(1-2\sin^2 \alpha\right) \\ &= -\frac{1}{3}\left( 1- \frac{2}{9} \right ) \\ &= -\frac{7}{27}. \ _\square \end{align}
If the roots of the quadratic equation $$8x^2+7x+a=0$$ are $$\sin \theta$$ and $$\cos 2\theta,$$ what is $$a?$$ From Vieta's formulas , we have \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} &\qquad (1) \\ \sin \theta \cos 2\theta &= \frac{a}{8}. &\qquad (2) \end{align} From $$(1),$$ we have \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} \\ 8(\sin \theta +1 - 2 \sin^2 \theta) + 7 &= 0 \\ (4\sin \theta +3)(4\sin \theta - 5) &= 0 \\ \Rightarrow \sin \theta &= -\frac{3}{4}, \frac{5}{4}. \end{align} Since $$\lvert \sin \theta \rvert \leq 1,$$ it follows that $$\sin \theta = -\frac{3}{4}.$$ Thus, we have \begin{align} \cos 2\theta &= 1-2\sin^2 \theta \\ &= 1-2\sin^2\left(-\frac{3}{4} \right) \\ &= -\frac{1}{8}. \end{align} Substituting this into $$(2)$$ gives \begin{align} \sin \theta \cos 2 \theta &= \frac{a}{8} \\ \left( -\frac{3}{4} \right) \left( -\frac{1}{8} \right ) &= \frac{a}{8} \\ \Rightarrow a&= \frac{3}{4}. \ _\square \end{align}

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## Solving Simple (to Medium-Hard) Trig Equations

Easy/Medium Hard

Solving trig equations use both the reference angles and trigonometric identities that you've memorized, together with a lot of the algebra you've learned. Be prepared to need to think in order to solve these equations.

In what follows, it is assumed that you have a good grasp of the trig-ratio values in the first quadrant , how the unit circle works, the relationship between radians and degrees , and what the various trig functions' curves look like, at least on the first period. If you're not sure of yourself, go back and review those topics first.

Content Continues Below

## Solve sin( x ) + 2 = 3 over the interval 0° ≤  x  < 360°

Just as with linear equations , I'll first isolate the variable-containing term:

sin( x ) + 2 = 3

sin( x ) = 1

Now I'll use the reference angles I've memorized to get my final answer.

So, in degrees, my answer is:

x = 90°

## Solve tan 2 (θ) + 3 = 0 on the interval 0° ≤ θ < 360°

There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution. I can see this when I slow down and do the steps. My first step is:

tan 2 (θ) = −3

Can any square (of a tangent, or of any other trig function) be negative ? No! So my answer is:

no solution

## Solve katex.render("\\boldsymbol{\\color{green}{\\small{ 2 \\cos^2(x) - \\sqrt{3\\,} \\cos(x) = 0 }}}", typed01); on the interval 0° ≤  x  < 360°

The left-hand side of this equation factors. I'm used to doing simple factoring like this:

2 y 2 + 3 y = 0

y (2 y + 3) = 0

...and then solving each of the factors. The same sort of thing works here. To solve the equation they've given me, I will start with the factoring:

I've done the algebra; that is, I've done the factoring and then I've solved each of the two factor-related equations. This created two trig equations. So now I can do the trig; namely, solving those two resulting trigonometric equations, using what I've memorized about the cosine wave. From the first equation, I get:

cos( x ) = 0:

x = 90°, 270°

From the second equation, I get:

2 cos( x ) = sqrt[3]:

x = 30°, 330°

Putting these the two solution sets together, I get the solution for the original equation as being:

x = 30°, 90°, 270°, 330°

## Solve sin 2 (θ) − sin(θ) = 2 on the interval 0 ≤ θ < 2π

First, I'll get everything over to one side of the "equals" sign:

sin 2 (θ) − sin(θ) − 2 = 0

This equation is "a quadratic in sine"; that is, the form of the equation is the quadratic-equation format:

a X 2 + b X + c = 0

In the case of the equation they're wanting me to solve, X = sin(θ) , a  = 1 , b  = −1 , and c  = −2 .

Since this is quadratic in form, I can apply some quadratic-equation methods. In the case of this equation, I can factor the quadratic:

( sin(θ) − 2 )( sin(θ) + 1 ) = 0

The first factor gives me the related trig equation:

sin(θ) = 2

But the sine is never more than 1 , so this equation is not solvable; it has no solution.

The other factor gives me the second related trig equation:

sin(θ) + 1 = 0

sin(θ) = −1

θ = (3/2)π

(If you're doing degrees-only solutions in your class, the solution value above equates to "270°".)

## Solve cos 2 (α) + cos(α) = sin 2 (α) on the interval 0° ≤ x < 360°

I can use a trig identity to get a quadratic in cosine:

cos 2 (α) + cos(α) = sin 2 (α)

cos 2 (α) + cos(α) = 1 − cos 2 (α)

2cos 2 (α) + cos(α) − 1 = 0

( 2cos(α) − 1 )( cos(α) + 1 ) = 0

cos(α) = 1/2, cos(α) = −1

α = 60°, 180°, 300°

## Solve sin(β) = sin(2β) on the interval 0° ≤ β < 360°

I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:

sin(β) = 2sin(β)cos(β)

sin(β) − 2sin(β) cos(β) = 0

sin(β) ( 1 − 2cos(β) ) = 0

sin(β) = 0,cos(β) = 1/2

The sine wave (from the first trig equation) is zero at 0° , 180° , and 360° . But, in the original exercise, 360° is not included, so this last solution value doesn't count, in this particular instance.

β = 0°, 60°, 180°, 300°

## Solve sin( x ) + cos( x ) = 1 on the interval 0° ≤ x < 360°

Hmm... I'm really not seeing anything here. It sure would have been nice if one of these trig expressions were squared...

Well, why don't I square both sides, and see what happens?

( sin( x ) + cos( x ) ) 2 = (1) 2

sin 2 ( x ) + 2sin( x )cos( x ) + cos 2 ( x ) = 1

[ sin 2 ( x + cos 2 ( x ) ] + 2sin( x )cos( x ) = 1

1 + 2sin( x )cos( x ) = 1

2sin( x )cos( x ) = 0

sin( x )cos( x ) = 0

Huh; go figure: I squared, and got something that I could work with. Nice!

From the last line above, either sine is zero or else cosine is zero, so my solution appears to be:

x = 0°, 90°, 180°, 270°

However (and this is important!), I squared to get this solution, and squaring is an "irreversible" process.

(Why? If you square something, you can't just square-root to get back to what you'd started with, because the squaring may have changed a sign somewhere.)

So, to be sure of my results, I need to check my answers in the original equation, to make sure that I didn't accidentally create solutions that don't actually count. Plugging back in, I see:

sin(0°) + cos(0°) = 0 + 1 = 1

...so the solution " x = 0° " works

sin(90°) + cos(90°) = 1 + 0 = 1

...so the solution " x = 90° " works, too

sin(180°) + cos(180°) = 0 + (−1) = −1

...oh, okay, so " x = 180° " does NOT work

sin(270°) + cos(270°) = (−1) + 0 = −1

...so " x = 270° " doesn't work, either

It's a good thing that I checked my solutions, because two of them don't actually work. They were created by the process of squaring.

My actual solution is:

x = 0°, 90°

Note: In the above, I could have stopped at this line:

...and used the double-angle identity for sine, in reverse, instead of dividing off the 2 in the next-to-last line in my computations. The answer would have been the same, but I would have needed to account for the solution interval:

2sin( x )cos( x ) = sin(2 x ) = 0

Then 2 x = 0°, 180°, 360°, 540° , etc, and dividing off the 2 from the x would give me x = 0°, 90°, 180°, 270° , which is the same almost-solution as before. After doing the necessary check (because of the squaring) and discarding the extraneous solutions, my final answer would have been the same as previously.

The squaring trick in the last example above doesn't come up often, but if nothing else is working, it might be worth a try. Keep it in mind for the next test.

URL: https://www.purplemath.com/modules/solvtrig.htm

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Chapter 5: Equations and Identities

## 5.2 Solving Equations

Algebra refresher.

Each of the following “solutions” contains an error. Find the error and supply a correct solution.

• \displaystyle [latex] \begin{aligned}[t] 3x^2 - 5x = 0 \\ 3x^2 = 5x\\ 3x = 5\\ x = \dfrac{5}{3}\\ \end{aligned}
• \displaystyle [latex] \begin{aligned}[t] 4x^2 - 5x =12 \\ x^2 = 3\\ x = \\\sqrt{3}\\ \end{aligned}
• \displaystyle [latex] \begin{aligned}[t] (2x+1)^2 = 4 \\ 2x+1 = 2\\ 2x = 1\\ x = \dfrac{1}{2}\\ \end{aligned}
• \displaystyle [latex] \begin{aligned}[t] 2x^2 - 4x = 3 \\ 2x(x-2) = 3\\ 2x = 3 \qquad x - 2 = 3\\ x = \dfrac{3}{2} \qquad x = 5\\ \end{aligned}

$\underline{\qquad\qquad\qquad\qquad}$

• We can't divide by $x.~~x = 0, \dfrac{5}{3}$
• 3 has two square roots. $~x = \pm\sqrt{3}$
• 4 has two square roots. $~x = \dfrac{-3}{2}, \dfrac{1}{2}$
• One side must be zero to use the zero-factor principle. No real solutions.

## Learning Objectives

• Use reference angles
• Solve equations by trial and error
• Use graphs to solve equations
• Solve trigonometric equations for exact values
• Use a calculator to solve trigonometric equations
• Solve trigonometric equations that involve factoring

## Introduction

It is important to distinguish between an algebraic expression and an equation . An equation is a statement that two algebraic expressions are equal. It may be true or false, depending on the values of any variables involved. Here are some examples of equations.

$5(2 + 6) = 5(2) + 5(6)$ $\sqrt{3^2} + \sqrt{4^2}= 3 + 4$ $x^2 + 3x = 10$

The first equation is true, the second is false, and the third equation is true only if $x = 2$ or $x = -5{.}$ When you solve an equation, you are finding the values of the variable that make the equation true.

## Example 5.19

• Evaluate $2x + \sqrt{x - 5}$ for $x = 9{.}$
• Solve $2x + \sqrt{x - 5} = 20{.}$
• Substitute $x = 9$ into the expression to find $2x + \sqrt{x - 5} = 2({9}) + \sqrt{{9} - 5}= 18 + \sqrt{4} = 20$
• We must find a value for $x$ that makes $2x + \sqrt{x - 5}$ equal to 20. In part (a) we saw that this expression equals 20 when $x = 9{,}$ so the solution of the equation $2x + \sqrt{x - 5} = 20$ is $x = 9{.}$

Checkpoint 5.20.

Try small integer values for $x{.}$

$x = 3$

You probably remember a number of algebraic techniques for solving equations of different types. Another useful equation-solving method uses graphs.

## Example 5.21.

Use a graph to solve the equation $x^3 - 2x^2 - 5x = -6{.}$

Checkpoint 5.22.

The graph does not cross the line $y = 0{.}$

## Trigonometric Equations

The first Ferris wheel was built for the Chicago World's Fair in 1893. It had a diameter of 250 feet and could carry 2160 people in 36 carriages. From the top of the wheel, passengers could see into four states. After loading all the passengers, the wheel made one revolution in nine minutes.

If you are in the bottom carriage of the Ferris wheel at the start of its revolution, your height after $t$ seconds is given by

$h = f(t) = 139 - 125\cos \left(\dfrac{2t}{3}\right)$

For how long are you more than 240 feet above the ground?

The figure below shows a graph of the height function and a horizontal line at $h = 240{.}$

## Solving Trigonometric Equations

In the example above, we used a graph to solve the equation $h = 240{,}$ or

$139 - 125\cos \left(\dfrac{2t}{3}\right) = 240$

To find a more precise solution, we can use algebraic methods. As an example, we'll solve the slightly simpler equation

$139 - 125\cos\theta = 240$

We'll look for all solutions for $\theta$ between $0°$ and $360°{.}$ We begin by isolating the trigonometric ratio on one side of the equation.

$139 - 125\cos \theta = 240$ Subtract 139 from both sides.} $- 125\cos \theta = 101$ Divide both sides by -125. $\cos\theta = -0.808$

We have solved equations like this one before: we use the inverse cosine to solve for $\theta{.}$ Remember that there are two angles between $0°$ and $360°$ that have a cosine of $-0.808{,}$ one in the second quadrant and one in the third quadrant. The calculator will give us only the second quadrant solution.

$\theta = \cos^{-1}(-0.808) = 143.9°$

To find the second solution, we need the third-quadrant angle whose cosine is $-0.808{.}$

Now, the reference angle for $143.9°$ is

$180° - 143.9° = 36.1°$

and the angle in the third quadrant with the same reference angle is

$180° + 36.1° = 216.1°$

(See the figure at right.) Thus, the other solution is $216.1°{.}$

To solve simple equations involving a single trigonometric ratio (either $\sin \theta, \cos \theta,$ or $\tan \theta$), we can follow the steps below.

## To Solve a Trigonometric Equation

1. Isolate the trigonometric ratio.

2. Find one solution in $0° \le\theta\le 360°$.

a. Give an exact solution if the trig ratio is one of the special values.

b. Otherwise use the inverse trig keys on a calculator.

3. Use reference angles to find a second solution (if there is one).

In step 3, note that if the first solution is a quadrantal angle, there may not be a second solution. Consider, for example, the equation $\cos \theta = -1{.}$

## Example 5.24.

Solve the equation $~~8 \sin \theta - 1 = 3~~$ for $0° \le\theta\le 360°{.}$

We begin by isolating the trig ratio.

$8 \sin \theta - 1 = 3$ Add 1 to both sides. $8 \sin \theta = 4$ Divide both sides by 8. $\sin \theta = \dfrac{4}{8} = \dfrac{1}{2}$

## Caution 5.25.

To solve the equation in the previous example, it is not enough to find the solution $\theta = 30°{;}$ we must find all the solutions between $0°$ and $360°{.}$ To find the solutions in other quadrants, we use reference angles.

Each intersection represents a solution of the equation. Thus, all of the angles coterminal with $30°$ and $150°$ are also solutions. We can easily find these solutions by adding integer multiples of $360°$ to $30°$ or $150°{.}$ This is why, when solving a trigonometric equation, we usually list only the solutions in one cycle, typically those between $0°$ and $360°{.}$

Checkpoint 5.27.

$\theta = 135°~ {and}~ \theta = 315°$

We can use a calculator to help us solve equations that do not involve special angles.

## Example 5.28.

First, we isolate the trig ratio.

$3 \tan \beta + 1 = -8\\ 3 \tan \beta = -9\\ \tan \beta = -3$

There are two angles with tangent -3, one in the second quadrant and one in the fourth quadrant. The calculator finds the solution

$\tan^{-1}(-3) = -71.565°$

This angle is in the fourth quadrant, but it is not between $0°$ and $360°{.}$

The angle we want is coterminal with $-71.565°{,}$ so we add $360°$ to get the first solution:

$B = \tan^{-1}(-3) + 360° = 288.435°$

The other solution is in the second quadrant, as shown in the figure. This angle is $180°$ less than the fourth quadrant solution:

$B = \tan^{-1}(-3) + 180° = 108.435°$

In the previous example, notice that the solutions are $180°$ apart. The solutions of an equation $\tan \theta = k$ always differ by $180°{,}$ so the use of reference angles is not necessary for these equations: once we have found one solution, we can add or subtract $180°$ to find the other solution between $0°$ and $360°{.}$ (See Homework Problems 69 and 70.)

Checkpoint 5.30.

$C = 41.410°~$ or $~ C = 318.590°$

Some trigonometric equations have no solution. As we can observe from their graphs or from their definitions, the sine and cosine functions only have values ranging from $-1$ to $1{.}$

## Example 5.31.

Solve $~~\dfrac{\sin \beta}{5} - 3 = 1~~$ for $0° \le\theta\le 360°{.}$

We begin by isolating $\sin \beta{.}$

$\dfrac{\sin \beta}{5} - 3 = 1$ Add 3 to both sides. $\dfrac{\sin \beta}{5} = 4$ Multiply both sides by 5. $\sin \beta = 20$

Because $\sin \beta$ is never greater than 1, there is no angle $\beta$ whose sine is 20. The equation has no solution.

Checkpoint 5.32.

No solution

## Graphical Solutions

We can use graphs to approximate the solutions to trigonometric equations.

## Example 5.33.

Use a graph to solve the equation $~~3\tan B + 1 = -8{.}$

Checkpoint 5.34.

## Equations with Squares of Trig Ratios

Simple quadratic equations can be solved by extracting roots. For example, to solve the equation

$4x^2 + 3 = 15$

we first isolate $x^2{:}$

$4x^2 = 12\\ x^2 = 3$

and then take square roots of both sides to find

$x = \pm \\\sqrt{3} \approx \pm 1.732$

Recall that a quadratic equation may have two real solutions, one (repeated) real solution, or no real solutions. We can use extraction of roots to solve trigonometric equations as well.

## Example 5.35.

We begin by isolating the power of tangent, $\tan^2 \theta{.}$

$4\tan^2 \theta = 12\\ \tan^2 \theta = 3$

Next, we solve for $\tan \theta$ by extracting roots.

$\tan \theta = \pm \\\sqrt{3}$

There are two angles between $0°$ and $360°$ with tangent $\\\sqrt{3}$ and two angles with tangent $-\\\sqrt{3}{,}$ making four solutions to this equation. We know that $\tan 60° = \\\sqrt{3}{,}$ so one of the solutions is $60°{.}$

$180° - 60° = 120° ~~ {and} ~~ 360° - 60° = 300°$

The four solutions are shown at right.

Checkpoint 5.36.

$\theta = 45°{,}$ $~\theta = 135°{,}$ $~\theta = 225°~{,}$ or $~\theta = 315°$

Other quadratic equations can be solved by factoring. For example, we can solve the equation

$4x^2 + 4x - 3 = 0$

by factoring the left side to get

$(2x + 3)(2x - 1) = 0$

Then we apply the Zero Factor Principle to set each factor equal to zero, and solve each equation.

$2x + 3 = 12 2x - 1= 0\\ x = \dfrac{-3}{2} x = \dfrac{1}{2}$

The solutions are $\dfrac{-3}{2}$ and $\dfrac{1}{2}{.}$

## Example 5.37.

Compare this equation to the algebraic equation $2x^2 - x - 1 = 0{.}$ Our equation has the same algebraic form, but with $x$ replaced by $\cos \theta {.}$

We can solve the algebraic equation by factoring $2x^2 - x - 1$ as $(2x + 1)(x - 1),$ and we'll use the same strategy on the trigonometric equation.

$2 \cos^2 \theta - \cos \theta - 1 = 0$ Factor the left side. $(2\cos \theta + 1)(\cos \theta - 1) = 0$ Set each factor equal to zero. $\cos \theta = \dfrac{-1}{2} ~~ { or } ~~\cos \theta = 1$

Now we solve each equation for $\theta{.}$ We know that $\cos 60° = \dfrac {1}{2},$ and the cosine function is negative in the second and third quadrants. The angles in those quadrants with reference angle $60°$ are

$180° - 60° = 120° ~~~{and} ~~~ 180° + 60° = 240°$

Also, $\cos \theta = 1$ when $\theta = 0°{.}$ Thus, the original equation has three solutions: $\theta = 120°, ~ \theta = 240°,$ and $\theta = 0°.$

Checkpoint 5.38.

$\theta = 42°{,}$ $~\theta = 138°{,}$ $~\theta = 199°~{,}$ or $~\theta = 341°$

## Snell's Law

When you view an object through a liquid, such as a spoon in a glass of water, or a fish in an aquarium, the object looks distorted or bent. This distortion is caused by refraction of light. Light rays bend when they pass from one medium to another, for instance from water to glass or from glass to air.

A light ray enters the boundary between the two media at a certain angle, called the angle of incidence , but leaves the boundary at a different angle, the angle of refraction . Both angles are acute angles measured from the normal line perpendicular to the boundary, as shown below.

The change of angle is caused by the fact that light travels at different speeds in different media. The relationship between the angle of incidence and the angle of refraction is given by Snell's Law:

$\dfrac{\sin \theta_1}{\sin \theta_2} = \dfrac{v_1}{v_2}$

where $\theta_1$ is the angle in the medium where light travels at speed $v_1{,}$ and $\theta_2$ is the angle where light travels at speed $v_2{.}$ The ratio of the speeds is called the index of refraction .

## Example 5.39.

The index of refraction from water to glass is 1.1. If light passes from water to glass with a $23°$ angle of incidence, what is the angle of refraction?

If $\theta$ is the angle of refraction, then from Snell's Law we have

$\dfrac{\sin 23°}{\sin \theta} = 1.1$ Multiply both sides by} \sin \theta. $\sin 23° = 1.1 \sin \theta$ Divide both sides by 1.1. $\sin \theta = \dfrac{\sin 23°}{1.1} = 0.3552$

Because $\sin \theta = 0.3552{,}$ $~ \theta = \sin^{-1}(0.3552) = 20.8°.$ (For Snell's Law we use only acute angles.) The angle of refraction is approximately $20.8°{.}$

## Checkpoint 5.40.

A light ray passes from water to glass with an $18°$ angle of incidence. What is the angle of refraction?

$16.315°$

## Section 5.2 Summary

• Zero Factor Principle
• Angle of incidence
• Angle of refraction

## Concepts

• An equation is a statement that two algebraic expressions are equal. It may be true or false.
• We can solve equations by trial and error, by using graphs, or by algebraic techniques.
• To solve a trigonometric equation, we first isolate the trigonometric ratio on one side of the equation.
• We use reference angles to find all the solutions between $0°$ and $360°{.}$
• We can use factoring or extraction of roots to solve some quadratic equations.

## Study Questions

• How many solutions between $0°$ and $360°$ does the equation $\cos \theta = k$ have for each value of $k$ between -1 and 0?
• How many solutions between $0°$ and $360°$ does the equation $\cos \theta = k$ have for each value of $k$ greater than 1?
• How many solutions between $0°$ and $360°$ does the equation $\sin^2 \theta = k$ have for each value of $k$ between -1 and 0?

Trigonometric Identities

Trigonometric Identities Problems & Solver Worksheet in PDF Format

## Squared Trig Identities [Squared Trigonometric Functions]

Squared Trig Identities , also known as Squared Trigonometric Functions, are fundamental mathematical relationships that involve squaring trigonometric functions or expressing them in terms of other trigonometric functions. These identities play a pivotal role in trigonometry, providing valuable tools for simplifying complex expressions, solving trigonometric equations, and gaining deeper insights into the behavior of trigonometric functions.

## Squared Trig Identities

Squared trigonometric identities involve squaring trigonometric functions or expressing squared trigonometric functions in terms of other trigonometric functions. These identities are useful in simplifying complex expressions and solving trigonometric equations.

Here are some fundamental squared trigonometric identities:

1.1. Pythagorean Identity: One of the most well-known squared trigonometric identities is the Pythagorean identity, which relates the squares of the sine and cosine functions. It states that for any angle θ: sin²θ + cos²θ = 1

1.2. Double Angle Formulas: The double angle formulas involve the squared trigonometric functions of angles that are double the given angle. For example: sin(2θ) = 2sinθcosθ cos(2θ) = cos²θ – sin²θ

1.3. Half Angle Formulas: The half angle formulas express the squared trigonometric functions of angles that are half the given angle. These formulas are useful in integration and solving trigonometric equations. For example: sin²(θ/2) = (1 – cosθ) / 2 cos²(θ/2) = (1 + cosθ) / 2

1.4. Sum and Difference Formulas: The sum and difference formulas for squared trigonometric functions express the square of the sum or difference of two angles in terms of the squares of the individual angles. For example: sin(α + β)² = sin²α + 2sinαsinβ + sin²β cos(α – β)² = cos²α + 2cosαcosβ + cos²β

## Identities of Squared Trigonometric Functions

Squared trigonometric functions are expressions where a trigonometric function is squared, such as sin²θ, cos²θ, tan²θ, etc. These identities are crucial in various fields, including physics, engineering, and calculus. Understanding these identities allows us to simplify trigonometric expressions, solve equations more efficiently, and gain insights into the behavior of trigonometric functions.

Here are some notable identities of squared trigonometric functions:

2.1. Reciprocal Identities: The reciprocal identities involve the squared trigonometric functions in the denominator. For example: csc²θ = 1/sin²θ sec²θ = 1/cos²θ cot²θ = 1/tan²θ

2.2. Sum and Difference Identities: Similar to squared trigonometric identities, sum and difference identities also exist for squared trigonometric functions. For example: sin²(α + β) = sin²α + 2sinαsinβ + sin²β cos²(α – β) = cos²α + 2cosαcosβ + cos²β

2.3. Double Angle Identities: Double angle identities for squared trigonometric functions help express the squares of double angles in terms of the original angle. For example: sin²(2θ) = 4sin²θcos²θ cos²(2θ) = 2cos²θ – 1

2.4. Half Angle Identities: The half-angle identities for squared trigonometric functions allow us to express the squares of half angles in terms of the original angle. For example: sin²(θ/2) = (1 – cosθ) / 2 cos²(θ/2) = (1 + cosθ) / 2

By mastering these identities, mathematicians, scientists, and engineers can manipulate trigonometric expressions with confidence and apply them to various problem-solving scenarios.

## How to Solve Trigonometric Equations with a Squared Function with the Square Root Property

Solving trigonometric equations involving squared functions is a common task in mathematics and physics. One effective method to solve such equations is by using the square root property. The square root property states that if x² = a , then x = ±√a , where ‘a’ is a constant.

Here’s a step-by-step guide on how to use the square root property to solve trigonometric equations with a squared function:

3.1. Isolate the squared function: Rewrite the given trigonometric equation in a way that you have a squared function on one side of the equation. For example, consider the equation sin²θ = 1/4 .

3.2. Apply the square root property: Take the square root of both sides of the equation. Don’t forget to consider both the positive and negative square roots. Using the example from step 1: sinθ = ±√(1/4) sinθ = ±1/2

3.3. Find all solutions: Trigonometric functions have periodicity, meaning they repeat after certain intervals (e.g., 360 degrees or 2π radians). To find all solutions, consider the general solutions for the trigonometric function involved. In our example, the solutions are: θ = 30° + 360°n θ = 150° + 360°n θ = 210° + 360°n θ = 330° + 360°n Where ‘n’ is an integer.

3.4. Check for extraneous solutions: After obtaining the solutions, check if they satisfy the original equation. In some cases, the solutions obtained may not be valid due to the domain restrictions of the squared trigonometric functions. Eliminate any solutions that do not satisfy the original equation.

By following these steps and practicing with various examples, you can confidently solve trigonometric equations involving squared functions using the square root property.

## Reciprocal, Quotient and Square Relations

Trigonometric functions have various interrelated properties that form the basis for solving complex trigonometric problems. Among these, reciprocal, quotient, and square relations are particularly important. Understanding these relations allows us to simplify trigonometric expressions, manipulate equations, and find connections between different trigonometric functions.

4.1. Reciprocal Relations: The reciprocal relations involve the reciprocal trigonometric functions, such as secant (sec), cosecant (csc), and cotangent (cot). These are defined as the reciprocals of cosine, sine, and tangent, respectively. The reciprocal relations are as follows: sec(θ) = 1 / cos(θ) csc(θ) = 1 / sin(θ) cot(θ) = 1 / tan(θ)

4.2. Quotient Relations: The quotient relations involve the tangent (tan) function and its reciprocal, cotangent (cot). The quotient relations are as follows: tan(θ) = sin(θ) / cos(θ) cot(θ) = cos(θ) / sin(θ)

4.3. Square Relations: Square relations involve squaring trigonometric functions or expressing squared trigonometric functions in terms of other trigonometric functions. For example, the Pythagorean identity is a square relation that relates the squares of sine and cosine: sin²(θ) + cos²(θ) = 1

4.4. Connection between Reciprocal and Quotient Relations: Reciprocal relations are related to quotient relations through the reciprocal identity. For example: csc(θ) = 1 / sin(θ) csc(θ) = 1 / (1 / cot(θ)) csc(θ) = cot(θ)

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## Trigonometry : Finding Trigonometric Roots

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : finding trigonometric roots.

We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

We then determine the angles that satisfy each solution within one revolution.

## Example Question #2 : Finding Trigonometric Roots

No solution exists

The equation now becomes:

So at what angles are the sine and cosine functions equal.  This occurs at

You may be wondering, "Why did you include

The reason is because once we substitute back the original variable, we will have to divide by 2.  This dividing by 2 will bring the last two answers within our range.

Dividing each answer by 2 gives us

## Example Question #8 : Finding Trigonometric Roots

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

## Example Question #9 : Finding Trigonometric Roots

No Solution

At this point, either use a unit circle diagram or a calculator to find the value.

## Report an issue with this question

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## 8.6: Solve Equations with Square Roots

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## Learning Objectives

By the end of this section, you will be able to:

• Use square roots in applications

Before you get started, take this readiness quiz.

• Simplify: ⓐ $$\sqrt{9}$$ ⓑ $$9^2$$. If you missed this problem, review Example 9.1.1 and Exercise 1.3.22 .
• Solve: 5(x+1)−4=3(2x−7). If you missed this problem, review Exercise 2.4.16 .
• Solve: $$n^2−6n+8=0$$. If you missed this problem, review Exercise 7.6.13 .

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

An equation in which the variable is in the radicand of a square root is called a radical equation .

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write $$\sqrt{a}$$ we mean the principal square root. So $$\sqrt{a} \ge 0$$ always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make $$\sqrt{a}$$ negative. This algebraic solution would not be a solution to the original radical equation ; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.

## Example $$\PageIndex{1}$$

For the equation $$\sqrt{x+2}=x$$:

• Is x=2 a solution?
• Is x=−1 a solution?

1. Is x=2 a solution?

2. Is x=−1 a solution?

## Example $$\PageIndex{2}$$

For the equation $$\sqrt{x+6}=x$$:

• Is x=−2 a solution?
• Is x=3 a solution?

## Example $$\PageIndex{3}$$

For the equation $$\sqrt{−x+2}=x$$:

• Is x=1 a solution?

For$$a \ge 0$$, $$(\sqrt{a})^2=a$$

## Example $$\PageIndex{4}$$

Solve: $$\sqrt{2x−1}=7$$

## Example $$\PageIndex{5}$$

Solve: $$\sqrt{3x−5}=5$$.

## Example $$\PageIndex{6}$$

Solve: $$\sqrt{4x+8}=6$$.

## Definition: SOLVE A RADICAL EQUATION.

• Isolate the radical on one side of the equation.
• Square both sides of the equation.
• Solve the new equation.

## Example $$\PageIndex{7}$$

Solve: $$\sqrt{5n−4}−9=0$$.

## Example $$\PageIndex{8}$$

Solve: $$\sqrt{3m+2}−5=0$$.

$$\frac{23}{3}$$

## Example $$\PageIndex{9}$$

Solve: $$\sqrt{10z+1}−2=0$$.

$$\frac{3}{10}$$

## Example $$\PageIndex{10}$$

Solve: $$\sqrt{3y+5}+2=5$$.

## Example $$\PageIndex{11}$$

Solve: $$\sqrt{3p+3}+3=5$$.

$$\frac{1}{3}$$

## Example $$\PageIndex{12}$$

Solve: $$\sqrt{5q+1}+4=6$$.

$$\frac{3}{5}$$

When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.

## Example $$\PageIndex{13}$$

Solve: $$\sqrt{9k−2}+1=0$$.

## Example $$\PageIndex{14}$$

Solve: $$\sqrt{2r−3}+5=0$$

no solution

## Example $$\PageIndex{15}$$

Solve: $$\sqrt{7s−3}+2=0$$.

## Definition: BINOMIAL SQUARES

$\begin{array}{cc} {(a+b)^2=a^2+2ab+b^2}&{(a−b)^2=a^2−2ab+b^2}\\ \nonumber \end{array}$

Don’t forget the middle term!

## Example $$\PageIndex{16}$$

Solve: $$\sqrt{p−1}+1=p$$.

## Example $$\PageIndex{17}$$

Solve: $$\sqrt{x−2}+2=x$$.

## Example $$\PageIndex{18}$$

Solve: $$\sqrt{y−5}+5=y$$.

## Example $$\PageIndex{19}$$

Solve: $$\sqrt{r+4}−r+2=0$$.

## Example $$\PageIndex{20}$$

Solve: $$\sqrt{m+9}−m+3=0$$.

## Example $$\PageIndex{21}$$

Solve: $$\sqrt{n+1}−n+1=0$$

When there is a coefficient in front of the radical, we must square it, too.

## Example $$\PageIndex{22}$$

Solve: $$3\sqrt{3x−5}−8=4$$.

## Example $$\PageIndex{23}$$

Solve: $$\sqrt{24a+2}−16=16$$.

$$\frac{127}{2}$$

## Example $$\PageIndex{24}$$

Solve: $$\sqrt{36b+3}−25=50$$.

$$\frac{311}{3}$$

## Example $$\PageIndex{25}$$

Solve: $$\sqrt{4z−3}=\sqrt{3z+2}$$.

## Example $$\PageIndex{26}$$

Solve: $$\sqrt{2x−5}=\sqrt{5x+3}$$.

## Example $$\PageIndex{27}$$

Solve: $$\sqrt{7y+1}=\sqrt{2y−5}$$.

Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.

## Example $$\PageIndex{28}$$

Solve: $$\sqrt{m}+1=\sqrt{m+9}$$.

## Example $$\PageIndex{29}$$

Solve: $$\sqrt{x}+3=\sqrt{x+5}$$.

## Example $$\PageIndex{30}$$

Solve: $$\sqrt{m}+5=\sqrt{m+16}$$.

## Example $$\PageIndex{31}$$

Solve: $$\sqrt{q−2}+3=\sqrt{4q+1}$$.

## Example $$\PageIndex{32}$$

Solve: $$\sqrt{y−3}+2=\sqrt{4y+2}$$.

no solution ​​​​​​​

## Example $$\PageIndex{33}$$

Solve: $$\sqrt{n−4}+5=\sqrt{3n+3}$$.

## Use Square Roots in Applications

As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.

We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.

## Definition: SOLVE APPLICATIONS WITH FORMULAS.

• Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Identify what we are looking for.
• Name what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

​​​​​​​ We used the formula A=L·W to find the area of a rectangle with length L and width W . A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is $$s^2$$ .

The formula $$A=s^2$$ gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s .

$\begin{array}{ll} {}&{A=s^2}\\ {\text{Take the square root of both sides.}}&{\sqrt{A}=\sqrt{s^2}}\\ {\text{Simplify.}}&{s=\sqrt{A}}\\ \nonumber \end{array}$

We can use the formula $$s=\sqrt{A}$$ to find the length of a side of a square for a given area.

## Definition: AREA OF A SQUARE

​​​​​​​ We will show an example of this in the next example.

## Example $$\PageIndex{34}$$

Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula $$s=\sqrt{A}$$ to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.

## Example $$\PageIndex{35}$$

Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. Use the formula $$s=\sqrt{A}$$ to find the length of each side of her lawn. Round your answer to the nearest tenth of a foot.

## Example $$\PageIndex{36}$$

Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. Use the formula $$s=\sqrt{A}$$ to find the length of each side of his mosaic. Round your answer to the nearest tenth of a foot.

​​​​​​​Another application of square roots has to do with gravity.

## Definition: FALLING OBJECTS

On Earth, if an object is dropped from a height of hh feet, the time in seconds it will take to reach the ground is found by using the formula,​​​​​​​

$$t=\frac{\sqrt{h}}{4}$$

​​​​​​​For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h=64 into the formula.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

## Example $$\PageIndex{37}$$

Christy dropped her sunglasses from a bridge 400 feet above a river. Use the formula $$t=\frac{\sqrt{h}}{4}$$ to find how many seconds it took for the sunglasses to reach the river.

## Exercise $$\PageIndex{38}$$

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula $$t=\frac{\sqrt{h}}{4}$$ to find how many seconds it took for the package to reach the ground.

## Example $$\PageIndex{39}$$

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula $$t=\frac{\sqrt{h}}{4}$$ to find how many seconds it took for the squeegee to reach the sidewalk.

3.5 seconds

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

## Definition: SKID MARKS AND SPEED OF A CAR

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula,

$$s=\sqrt{24d}$$​​​​​​​

## Example $$\PageIndex{40}$$

After a car accident, the skid marks for one car measured 190 feet. Use the formula $$s=\sqrt{24d}$$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

## Example $$\PageIndex{41}$$

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula $$s=\sqrt{24d}$$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

## Example $$\PageIndex{42}$$

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula $$s=\sqrt{24d}$$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

## Key Concepts

• Check the answer. Some solutions obtained may not work in the original equation.

• On Earth, if an object is dropped from a height of hh feet, the time in seconds it will take to reach the ground is found by using the formula$$t=\frac{\sqrt{h}}{4}$$.
• If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula $$s=\sqrt{24d}$$.

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## 3.2.7: Trigonometric Equations Using the Quadratic Formula

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The quadratic formula with a trigonometric function in place of the variable.

Solving equations is a fundamental part of mathematics. Being able to find which values of a variable fit an equation allows us to determine all sorts of interesting behavior, both in math and in the sciences. Solving trig equations for angles that satisfy the equation is one application of mathematical methods for solving equations. Suppose someone gave you the following equation:

$$3 \sin^2 \theta +8 \sin\theta −3=0$$

## Quadratic Functions with Trigonometric Equations

When solving quadratic equations that do not factor, the quadratic formula is often used.

Remember that the quadratic equation is:

$$ax^2+bx+c=0$$ (where $$a$$, $$b$$, and $$c$$ are constants)

In this situation, you can use the quadratic formula to find out what values of "x" satisfy the equation.

The same method can be applied when solving trigonometric equations that do not factor. The values for $$a$$ is the numerical coefficient of the function's squared term, $$b$$ is the numerical coefficient of the function term that is to the first power and $$c$$ is a constant. The formula will result in two answers and both will have to be evaluated within the designated interval.

## Solving for Unknown Values

1. Solve $$3 \cot ^2 x−3 \cot x=1$$ for exact values of $$x$$ over the interval $$[0,2\pi ]$$.

\begin{aligned} 3\cot ^2 x−3\cot x &=1 \\ 3\cot ^2x−3\cot x−1&=0 \end{aligned}

The equation will not factor. Use the quadratic formula for $$\cot x$$, $$a=3$$, $$b=−3$$, $$c=−1$$.

\begin{aligned} \cot x&=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\ \cot x&=\dfrac{-(-3) \pm \sqrt{(-3)^{2}-4(3)(-1)}}{2(3)}\\ \cot x&=\dfrac{3 \pm \sqrt{9+12}}{6}\\ \cot x&=\dfrac{3+\sqrt{21}}{6} &\quad \text { or } \quad& \cot x=\dfrac{3-\sqrt{21}}{6}\\ \cot x&=\dfrac{3+4.5826}{6} &\quad& \cot x=\dfrac{3-4.5826}{6}\\ \cot x&=1.2638 &\quad &\cot x=-0.2638\\ \tan x&=\dfrac{1}{1.2638} &\quad &\tan x=\dfrac{1}{-0.2638}\\ x&=0.6694,3.81099 & &x=1.8287,4.9703 \end{aligned}

2. Solve $$−5 \cos ^2 x+9 \sin x+3=0$$ for values of x over the interval $$[0,2\pi ]$$.

Change $$\cos ^2 x$$ to $$1−\sin ^2 x$$ from the Pythagorean Identity.

\begin{aligned} −5\cos ^2 x +9\sin x+3 &=0 \\ −5(1−\sin ^2 x )+9\sin x+3 &=0 \\ −5+5\sin ^2 x +9\sin x+3 &=0 \\ 5\sin ^2 x +9\sin x−2 &=0 \end{aligned}

$$\begin{array}{l} \sin x=\dfrac{-9 \pm \sqrt{9^{2}-4(5)(-2)}}{2(5)} \\ \sin x=\dfrac{-9 \pm \sqrt{81+40}}{10} \\ \sin x=\dfrac{-9 \pm \sqrt{121}}{10} \\ \sin x=\dfrac{-9+11}{10} \text { and } \sin x=\dfrac{-9-11}{10} \\ \sin x=\dfrac{1}{5} \text { and }-2 \\ \sin ^{-1}(0.2) \text { and } \sin ^{-1}(-2) \end{array}$$

$$x\approx .201 \text{rad}$$ and $$\pi −.201\approx 2.941$$

This is the only solutions for $$x$$ since $$−2$$ is not in the range of values.

3. Solve $$3\sin ^2 x −6\sin x−2=0$$ for values of $$x$$ over the interval $$[0,2\pi ]$$.

$$\begin{array}{l} \quad 3 \sin ^{2} x-6 \sin x-2=0 \\ \sin x=\dfrac{6 \pm \sqrt{(-6)^{2}-4(3)(-2)}}{2(3)} \\ \sin x=\dfrac{6 \pm \sqrt{36-24}}{6} \\ \sin x=\dfrac{6 \pm \sqrt{12}}{6} \\ \sin x=\dfrac{6+3.46}{10} \text { and } \sin x=\dfrac{6-3.46}{10} \\ \sin x=.946 \text { and } .254 \\ \sin ^{-1}(0.946) \text { and } \sin ^{-1}(0.254) \end{array}$$

$$x\approx 71.08 \text{ deg}$$ and $$\approx 14.71 \text{ deg}$$

## Example $$\PageIndex{1}$$

Earlier, you were asked to solve an equation.

The original equation to solve was:

3sin2\theta +8sin\theta −3=0

Using the quadratic formula, with a=3,b=8,c=−3, we get:

$$\sin\theta =\dfrac{−b \pm \sqrt{b^2−4ac}}{2a}=\dfrac{−8\pm \sqrt{64−(4)(3)(−3)}}{6}=\dfrac{−8\pm \sqrt{100}}{6}=\dfrac{−8\pm 10}{6}=\dfrac{1}{3} \text{or} −3$$

The solution of -3 is ignored because sine can't take that value, however:

$$\sin^{−1} \dfrac{1}{3}=19.471^{\circ}$$

## Example $$\PageIndex{2}$$

Solve $$\sin ^2 x −2\sin x−3=0$$ for $$x$$ over $$[0,\pi ]$$.

You can factor this one like a quadratic.

\begin{aligned} \sin ^{2} x-2 \sin x-3 &=0 & & \\ (\sin x-3)(\sin x+1) &=0 & & \\ \sin x-3=0 & & & \\ \sin x=3 & & \text { or } & \sin x+1=0 \\ x=\sin ^{-1}(3) & & &x=\dfrac{3 \pi}{2} \end{aligned}

For this problem the only solution is $$\dfrac{3\pi}{2}$$ because sine cannot be $$3$$ (it is not in the range).

## Example $$\PageIndex{3}$$

Solve $$\tan^2 x+\tan x−2=0$$ for values of $$x$$ over the interval $$\left[−\dfrac{\pi}{2},\; \dfrac{\pi}{2}\right]$$.

$$\tan^2 x+\tan x−2=0$$

\begin{aligned} -1 \pm \sqrt{1^{2}-4(1)(-2)} &=\tan x \\ 2 & \\ \dfrac{-1 \pm \sqrt{1+8}}{2} &=\tan x \\ \dfrac{-1 \pm 3}{2} &=\tan x \\ \tan x &=-2 \quad \text { or } \; 1 \end{aligned}

$$\tan x=1$$ when $$x=\dfrac{\pi}{4}$$, in the interval $$\left[−\dfrac{\pi}{2},\; \dfrac{\pi}{2}\right]$$

$$\tan x=−2$$ when $$x=−1.107 \text{rad}$$

## Example $$\PageIndex{4}$$

Solve the trigonometric equation such that $$5 \cos^2 \theta −6 \sin\theta =0$$ over the interval $$[0,2\pi ]$$.

$$5\cos^2\theta −6\sin \theta =0$$ over the interval $$[0,2\pi ]$$.

\begin{aligned} 5\left(1-\sin ^{2} x\right)-6 \sin x &=0 \\ -5 \sin ^{2} x-6 \sin x+5 &=0 \\ 5 \sin ^{2} x+6 \sin x-5 &=0 \\ -6 \pm \sqrt{6^{2}-4(5)(-5)} &=\sin x \\ 2(5) & \\ \dfrac{-6 \pm \sqrt{36+100}}{10} &=\sin x \\ \dfrac{-6 \pm \sqrt{136}}{10} &=\sin x \\ \dfrac{-6 \pm 2 \sqrt{34}}{10} &=\sin x \\ \dfrac{-3 \pm \sqrt{34}}{5} &=\sin x \end{aligned}

$$x=\sin^{−1}\left(\dfrac{−3+\sqrt{34}}{5}\right)$$ or $$\sin^{−1}\left(\dfrac{−3−\sqrt{34}}{5}\right) x=0.6018 \text{ rad}$$ or $$2.5398 \text{ rad}$$ from the first expression, the second expression will not yield any answers because it is out the range of sine.

Solve each equation using the quadratic formula.

• $$3x^2+10x+2=0$$
• $$5x^2+10x+2=0$$
• $$2x^2+6x−5=0$$

Use the quadratic formula to solve each quadratic equation over the interval $$[0,2\pi )$$.

• $$3\cos^2(x)+10\cos(x)+2=0$$
• $$5\sin^2(x)+10\sin(x) +2=0$$
• $$2\sin^2(x)+6\sin(x) −5=0$$
• $$6\cos^2(x)−5\cos(x)−21=0$$
• $$9\tan^2(x)−42\tan(x) +49=0$$
• $$\sin^2(x)+3\sin(x) =5$$
• $$3\cos^2(x)−4\sin(x) =0$$
• $$−2\cos^2(x)+4\sin(x) =0$$
• $$\tan^2(x)+\tan(x) =3$$
• $$\cot^2(x)+5\tan(x) +14=0$$
• $$\sin^2(x)+\sin(x) =1$$
• What type of sine or cosine equations have no solution?

Video: Solving Trigonometric Equations Using the Quadratic Formula

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3. Solve trigonometric equations by taking square root

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