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SECTION 9.5 SOLVING TRIGONOMETRIC EQUATIONS 739
LEARNING OBJECTIVES
In this section, you will:
ō Solve linear trigonometric equations in sine and cosine.
ō Solve equations involving a single trigonometric function.
ō Solve trigonometric equations using a calculator.
ō Solve trigonometric equations that are quadratic in form.
ō Solve trigonometric equations using fundamental identities.
ō Solve trigonometric equations with multiple angles.
ō Solve right triangle problems.
9.5 SOLVING TRIGONOMETRIC EQUATIONS
Figure 1 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 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 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 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.
740 CHAPTER 9 TRIGONOMETRIC IDENTITIES AND EQUATIONS
Example 1 Solving a Linear Trigonometric Equation Involving the Cosine Function
1
__
Find all possible exact solutions for the equation cos θ =! ! .
2
Solution From the unit circle, we know that 1
__
cos θ =! !
2
π 5π
__ ___
θ = , 3
3
These are the solutions in the interval [0, 2π]. All possible solutions are given by
π 5π
__ ___
± 2kπ
θ = ± 2kπ and θ = 3
where k is an integer. 3
Example 2 Solving a Linear Equation Involving the Sine Function
1
__
Find all possible exact solutions for the equation sin t =! ! .
2
Solution Solving for all possible values of t means that solutions include angles beyond the period of 2π. From Section
π 5π
__ ___
. But the problem is asking for all possible values that
9.2 Figure 2, we can see that the solutions are t = and t = 6
solve the equation. Therefore, the answer is 6
π 5π
__ ___
± 2πk
± 2πk and t =
t = 6 6
where k is an integer.
H!w To…
Given a trigonometric equation, solve using algebra.
1. Look for a pattern that suggests an algebraic property, such as the difference of squares 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 resulting expressions.
5. Solve for the angle.
Example 3 Solve the Trigonometric Equation in Linear Form
Solve the equation exactly: 2cos θ − 3 = − 5, 0 ≤ θ < 2π.
Solution Use algebraic techniques to solve the equation.
2cos θ − 3 = −5
2cos θ = −2
cos θ = −1
θ = π
Try It #1
Solve exactly the following linear equation on the interval [0, 2π): 2sin 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 Section 9.2 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, not 2π.
π
__
Further, the domain of tangent is all real numbers with the exception of odd integer multiples of 2 , unless, of course,
a problem places its own restrictions on the domain.
SECTION 9.5 SOLVING TRIGONOMETRIC EQUATIONS 741
Example 4 Solving a Problem Involving a Single Trigonometric Function
2
Solve the problem exactly: 2sin θ − 1 = 0, 0 ≤ θ < 2π.
Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to
isolate sin θ. Then we will find the angles.
2
2sin θ − 1 = 0
2
2sin θ = 1
1
2 __
sin θ =! !
2 __
— 1
2 __
sin θ = ±
√
√2
—
1 √2
_ ____
= ± !
sin θ = ± !
— 2
2
√
π 3π 5π 7π
__ ___ ___ ___
, ,
θ = 4 , 4 4 4
Example 5 Solving a Trigonometric Equation Involving Cosecant
Solve the following equation exactly: csc θ = −2, 0 ≤ θ < 4π. (Note the domain)
Solution We want all values of θ for which csc θ = −2 over the interval 0 ≤ θ < 4π.
csc θ = −2
1
____ = −2
(Add 2π to the previous answers!)
sin θ
1
__
sin θ = − !
2
7π 11π 19π 23π
___ ___ ___ ___
θ = , , ,
6 6 6 6
1
_
Analysis As sin θ = − ! , notice that all four solutions are in the third and fourth quadrants.
2
Example 6 Solving an Equation Involving Tangent
π
__
Solve the equation exactly: tan θ − = 1, 0 ≤ θ < 2π.
( 2 )
π
__
Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of 4 , the tangent has a
π π
__ __
value of 1. However, the angle we want is θ − . Thus, if tan = 1, then
( 2 ) ( 4 )
π π
__ __
θ − =
2 4
3π
___
± kπ
θ = 4
Over the interval [0, 2π), we have two solutions:
3π 3π 7π
___ ___ ___
and θ = + π =
θ = 4 4 4
Try It #2
—
Find all solutions for tan x = 3 .
√
Example 7 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.
742 CHAPTER 9 TRIGONOMETRIC IDENTITIES AND EQUATIONS
Solution We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign.
2(tan x) + 2(3) = 5 + tan x
2tan x + 6 = 5 + tan x
2tan x − tan x = 5 − 6
tan x = −1
3π 7π
__ __
and θ = .
There are two angles on the unit circle that have a tangent value of −1: θ = 4 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.
Example 8 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.
Solution Make sure mode is set to radians. To find θ, use the inverse sine function. On most calculators, you will
ND −1
need to push the 2 button and then the SIN button to bring up the sin function. What is shown on the screen is
−1 −1
sin ( . The calculator is ready for the input within the parentheses. For this problem, we enter sin (0.8), and press
ENTER. Thus, to four decimals places,
−1
sin (0.8) ≈ 0.9273
The solution is
θ ≈ 0.9273 ± 2πk
The angle measurement in degrees is
θ ≈ 53.1°
θ ≈ 180° − 53.1°
≈ 126.9°
Analysis 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.
Example 9 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.
Solution We can begin with some algebra.
sec θ = −4
1
____ = −4
cos θ
1
__
cos θ = !
−
4
Check that the MODE is in radians. Now use the inverse cosine function.
1
−1 __
cos − ! ≈ 1.8235
( 4 )
θ ≈ 1.8235 + 2πk
π
__
≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is in quadrant II.
Since 2
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.
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