Solving Trigonometric Equations

Solving Linear Trigonometric Equations in Sine and CosineTrigonometric equations are, as the name implies, equations that involve trigonometricfunctions.. Solving a Linear Trigonometric

Solving Trigonometric

Equations

By:

OpenStaxCollege

Egyptian pyramids standing near a modern city (credit: Oisin Mulvihill)

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 fractalgeometry, engineering, and architecture Often, the angle of elevation and the angle ofdepression are found using similar triangles

In earlier sections of this chapter, we looked at trigonometric identities Identities aretrue for all values in the domain of the variable In this section, we begin our study oftrigonometric 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 trigonometricfunctions 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 asoften, we will be asked to find all possible solutions, and as trigonometric functionsare periodic, solutions are repeated within each period In other words, trigonometricequations may have an infinite number of solutions Additionally, like rationalequations, the domain of the function must be considered before we assume that anysolution is valid The period of both the sine function and the cosine function is 2π In

other words, every 2π units, the y-values repeat If we need to find all possible solutions, then we must add 2πk, where 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 2π :

sin θ = sin(θ ± 2kπ)

There are similar rules for indicating all possible solutions for the other trigonometricfunctions Solving trigonometric equations requires the same techniques as solvingalgebraic equations We read the equation from left to right, horizontally, like asentence We look for known patterns, factor, find common denominators, andsubstitute certain expressions with a variable to make solving a more straightforwardprocess However, with trigonometric equations, we also have the advantage of usingthe 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 = 12

From the unit circle, we know that

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = 12

Solving for all possible values of t means that solutions include angles beyond the period

of 2π From the[link], we can see that the solutions are t = π6 and t = 5π6 But the problem

is asking for all possible values that solve the equation Therefore, the answer is

t = π6 ± 2πk and t = 5π6 ± 2πk

where k is an integer.

How To

Given a trigonometric equation, solve using algebra.

1 Look for a pattern that suggests an algebraic property, such as the difference ofsquares or a factoring opportunity

2 Substitute the trigonometric expression with a single variable, such as x or u.

3 Solve the equation the same way an algebraic equation would be solved

4 Substitute the trigonometric expression back in for the variable in the resultingexpressions

5 Solve for the angle

Solve the Trigonometric Equation in Linear Form

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

Use algebraic techniques to solve the equation

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 [link]) Weneed to make several considerations when the equation involves trigonometric functionsother than sine and cosine Problems involving the reciprocals of the primarytrigonometric functions need to be viewed from an algebraic perspective In otherwords, 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, not 2π Further,the domain of tangent is all real numbers with the exception of odd integer multiples of

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 sin2θ − 1 = 0, 0 ≤ θ < 2π

As this problem is not easily factored, we will solve using the square root property First,

we use algebra to isolate sinθ Then we will find the angles

√2 = ± √2

2

θ = π4, 3π4 , 5π4 , 7π4

Solving a Trigonometric Equation Involving Cosecant

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

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

Solve the equation exactly: tan(θ − π2)= 1, 0 ≤ θ < 2π.

Recall that the tangent function has a period of π On the interval[0, π), and at theangle of π4, the tangent has a value of 1 However, the angle we want is(θ − π2) Thus, iftan(π

Identify all Solutions to the Equation Involving Tangent

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

We can solve this equation using only algebra Isolate the expression tan x on the left

side of the equals sign

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, where θ is in radians

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

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.Using a Calculator to Solve a Trigonometric Equation Involving Secant

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

We can begin with some algebra

cos− 1( − 14) ≈ 1.8235

θ ≈ 1.8235 + 2πk

Since π2 ≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is

in quadrant II Cosine is also negative in quadrant III Note that a calculator will onlyreturn an angle in quadrants I or II for the cosine function, since that is the range of theinverse cosine See[link]

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 The other solution in quadrant III is

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can usealgebra 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 andone of the terms is squared, think about the standard form of a quadratic Replace the

trigonometric function with a variable such as x or 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: cos2θ + 3 cos θ − 1 = 0, 0 ≤ θ < 2π.

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

This terminal side of the angle lies in quadrant I Since cosine is also positive in quadrant

IV, the second solution is

Solve the equation exactly: 2 sin2θ − 5 sin θ + 3 = 0, 0 ≤ θ ≤ 2π.

Using grouping, this quadratic can be factored Either make the real substitution,

sin θ = u, or imagine it, as we factor:

The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, 7π6 , 11π6

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 [link] On the interval 0 ≤ θ < 2π, 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 thatare 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[link] as well

Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2 sin2θ − 3 sin θ + 1 = 0, 0 ≤ θ < 2π

We can factor using grouping Solution values of θ can be found on the unit circle:(2 sin θ − 1)(sin θ − 1) = 0

2 sin θ − 1 = 0

sin θ = 12

θ = π6, 5π6sin θ = 1

θ = π2Try It

Solve the quadratic equation 2 cos2θ + cos θ = 0

2, 2π3 , 4π3 , 3π2

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can alsouse the fundamental identities because they make solving equations simpler Rememberthat 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 tomatch 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 < 2π.

cos x cos(2x) + sin x sin(2x) = √3

2Notice that the left side of the equation is the difference formula for cosine

cos x cos(2x) + sin x sin(2x) = √3

From the unit circle in[link], we see that cos x = √23 when x = π6, 11π6

Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: 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 angle identity involving only cosine:

double-cos(2θ) = cos θ2cos2θ − 1 = cos θ

2 cos2θ − cos θ − 1 = 0

(2 cos θ + 1)(cos θ − 1) = 0

2 cos θ + 1 = 0cos θ = − 12cos θ − 1 = 0cos θ = 1

So, if cos θ = − 12, then θ = 2π3 ± 2πk and θ = 4π3 ± 2πk; if cos θ = 1, then θ = 0 ± 2πk.

Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin2θ, 0 ≤ θ < 2π

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

cos θ = − 1

θ = πOur solutions are θ = 2π3 , 4π3, π

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have amultiple angle, such as sin(2x)or cos(3x) When confronted with these equations, recall

that y = sin(2x)is a horizontal compression by a factor of 2 of the function y = sin x On

an interval of 2π, we can graph two periods of y = sin(2x), as opposed to one cycle of

y = sin x This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x) = 0 compared to sin x = 0 This information will help

us solve the equation

Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos(2x) = 12on[0, 2π)

We can see that this equation is the standard equation with a multiple of an angle

If cos(α) = 12, we know α is in quadrants I and IV While θ = cos− 1 12 will only yieldsolutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 12will be in quadrants I and IV

Therefore, the possible angles are θ = π3and θ = 5π3 So, 2x = π3 or 2x = 5π3, which means

that x = π6 or x = 5π6 Does this make sense? Yes, because cos(2(π

6) ) = cos(π

3)= 12.Are there any other possible answers? Let us return to our first step

In quadrant I, 2x = π3, so x = π6 as noted Let us revolve around the circle again:

2x = π3 + 4π

= π3 + 12π3

= 13π3

x = 13π6 > 2π, so this value for x is larger than 2π, so it is not a solution on[0, 2π)

In quadrant IV, 2x = 5π3, so x = 5π6 as noted Let us revolve around the circle again:

x = 17π6 > 2π, so this value for x is larger than 2π, so it is not a solution on[0, 2π)

Our solutions are x = π6, 5π6, 7π6, and 11π6 Note that whenever we solve a problem in theform of sin(nx) = c, we must go around the unit circle n times.

Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involveapplying the properties of right triangles and the Pythagorean Theorem We begin

with the familiar Pythagorean Theorem, a2+ b2 = c2, and model an equation to fit asituation

Using the Pythagorean Theorem to Model an Equation

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

One of the cables that anchors the center of the London Eye Ferris wheel to the groundmust 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[link]

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

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 wallfor every 4 feet of ladder length Find the angle that a ladder of any length forms withthe 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 4a feet See[link]

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

cos θ = a

4a =

14cos− 1(1

4) ≈ 75.5∘

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

• Solving Trigonometric Equations I

• Solving Trigonometric Equations II

• Solving Trigonometric Equations III

• Solving Trigonometric Equations IV

• Solving Trigonometric Equations V

• Solving Trigonometric Equations VI

Key Concepts

• When solving linear trigonometric equations, we can use algebraic techniquesjust as we do solving algebraic equations Look for patterns, like the difference

of squares, quadratic form, or an expression that lends itself well to

substitution See[link],[link], and[link]

• Equations involving a single trigonometric function can be solved or verifiedusing the unit circle See[link],[link], and[link], and[link]

• We can also solve trigonometric equations using a graphing calculator See[link]and[link]

• Many equations appear quadratic in form We can use substitution to make theequation appear simpler, and then use the same techniques we use solving analgebraic quadratic: factoring, the quadratic formula, etc See[link],[link],[link], and[link]

• We can also use the identities to solve trigonometric equation See[link],

[link], and[link]

• We can use substitution to solve a multiple-angle trigonometric equation,

which is a compression of a standard trigonometric function We will need totake the compression into account and verify that we have found all solutions

on the given interval See[link]