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Trigonometry Through Wayfinding and Navigation Across the Pacific

Kamuela E Yong

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Section 3.5 Basic Trigonometric Equations

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Trigonometric equations are equations involving trigonometric functions such as sine, cosine, and tangent. These equations seek to find specific values, known as solutions, that satisfy the equation.
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Trigonometric functions are inherently periodic, meaning they repeat their values at regular intervals. As a result, trigonometric equations may have multiple solutions due to this periodic nature. In fact, some equations may have infinitely many solutions.
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To address all possible solutions, we use a technique known as a general solution. This method involves initially identifying solutions within a single period of the trigonometric function. We then extend these solutions by adding integer multiples of the period of the trigonometric function.
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In this section, we will explore various techniques for effectively solving trigonometric equations, including methods for finding general solutions.
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Subsection 3.5.1 Solving Equations with a Single Trigonometric Function

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Example 3.5.1.

Solve the equation
sin⁡θ=12.

Solution.

To solve the equation sin⁡θ=12, our initial instinct might lead us to take the inverse sine of both sides:
sin−1⁡(sin⁡θ)=sin−1⁡(12)
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resulting in
θ=π6.
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While this is a valid solution, it’s important to recognize that there are additional solutions to consider.
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Since the sine function is positive in both Quadrant I and Quadrant II, we can find another solution in Quadrant II by using the reference angle π6 and the methods described in Subsection 1.5.2. Therefore, the equivalent angle in Quadrant II is θ=π−π6=5π6.
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However, these two angles, θ=π6 or θ=5π6, are not the only solutions. Recall that the sine function has a period of 2π, meaning that adding or subtracting any integer multiples of 2π to these angles will also give you solutions. For example, θ=π6+4π and θ=5π6−10π are both solutions.
The graph of y=sin(x) crosses the y-value of 1/2 multiple times.
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Thus, the general solution to sin⁡θ=12 can be expressed as:
θ=π6+2kπorθ=5π6+2kπ,
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where k is any integer.
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Example 3.5.2.

Solve the equation 2cos⁡θ+2=0, list six solutions.

Solution.

First we will isolate cos⁡θ.
2cos⁡θ+2=02cos⁡θ=−2cos⁡θ=−22.
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Thus we have the general solution θ=3π4+2kπ or θ=5π4+2kπ for any integer k.
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To get specific solutions, we select specific values of k:
k=−1:θ=3π4+2(−1)π=−5π4,θ=5π4+2(−1)π=−3π4k=0:θ=3π4+2(0)π=3π4,θ=5π4+2(0)π=5π4k=1:θ=3π4+2(1)π=11π4,θ=5π4+2(1)π=13π4
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Subsection 3.5.2 Solving Trigonometric Equations with Square Terms

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Example 3.5.3.

Solve cos2⁡θ=12 where 0≤θ<2π on the interval 0≤θ<2π.

Solution.

We will first solve for cos⁡θ. We begin by taking the square root of both sides of the equation and simplify:
cos2⁡θ=±12cos⁡θ±12=±22
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From Table 1.5.18, when cos⁡θ=22, we have θ=π4 or θ=7π4; and when cos⁡θ=−22, we have θ=3π4 or θ=5π4.
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Thus, our solutions are:
θ=π4,3π4,5π4,7π4.
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Subsection 3.5.3 Solving Trigonometric Equations by Factoring

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Example 3.5.4.

Solve cos2⁡x−4cos⁡x+3=0.

Solution.

To solve this equation, let’s make a substitution to simplify it. We’ll let y=cos⁡x, so the equation becomes y2−4y+3=0.
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Factoring the quadratic equation, we obtain (y−3)(y−1)=0. Setting each factor equal to zero, we find two potential solutions: y=3 or y=1. We are not done because we need to solve for x and not y.
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Substituting back cos⁡x for y, we find that cos⁡x=3 is not a valid solution, as the range of cosine is limited to [-1, 1]. However, cos⁡x=1 yields a solution of x=0 for one period.
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Therefore, the general solution to the equation is:
x=0+2kπ=2kπ,
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where k is any integer.
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Example 3.5.5.

Solve 2cos⁡θsin⁡θ+3cos⁡θ=0.

Solution.

We begin by factoring cos⁡θ:
2cos⁡θsin⁡θ+3cos⁡θ=0cos⁡θ(2sin⁡θ+3)=0
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Thus we get two equations: cos⁡θ=0 and 2sin⁡θ+3=0.
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From cos⁡θ=0 we get θ=π2 or θ=3π2.
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From the second equation, we isolate sin⁡θ:
2sin⁡θ+3=02sin⁡θ=−3sin⁡θ=−32
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Thus we get θ=4π3 or θ=5π3.
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We get the general solutions by adding integer multiples of 2π to get
θ=π2+2kπ,θ=3π2+2kπ,θ=4π3+2kπ,θ=5π3+2kπ
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where k is any integer.
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Subsection 3.5.4 Solving a Trigonometric Equation with a Calculator

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Example 3.5.6.

Use a calculator to solve 3tan⁡θ=2 on the interval 0≤θ<2π. Express your answer in radians, rounded to two decimals.

Solution.

We begin by isolating tan⁡θ:
tan⁡θ=23
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Next, we take the inverse tangent and use a calculator to obtain
θ=tan−1⁡(23)≈0.588002603548
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Rounding to two decimals, we get θ=0.59 radians, which is in Quadrant I since 0<0.59<π2. Another Quadrant where tan⁡θ=23 is in Quadrant III. Using the methods in Subsection 1.5.2 we get the other angle: θ=0.59+π.
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Exercises 3.5.5 Exercises

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Exercise Group.

Solve each equation on the interval 0≤θ<2π
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Exercise Group.

Solve each equation, giving the general formula for each solution. List six specific solutions.
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13.
sin⁡θ=−22
Answer.
5π4+2kπ, 7π4+2kπ; −3π4, −π4, 5π4, 7π4, 13π4, 15π4
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14.
cos⁡θ=12
Answer.
π3+2kπ, 5π3+2kπ; −5π3, −π3, π3, 5π3, 7π3, 11π3
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15.
tan⁡θ=−1
Answer.
π4+kπ; −3π4, π4, 5π4, 9π4, 13π4, 17π4
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16.
cot⁡θ=−3
Answer.
5π6+kπ, 11π6+kπ; −π6, 5π6, 11π6, 17π6, 23π6, 29π6
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Exercise Group.

Solve each equation on the interval 0≤θ<2π. Express your answer in radians, rounded to two decimals.
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Exercise Group.

Solve each equation on the interval 0≤θ<2π.
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27.
(2cos⁡θ−1)(csc⁡θ+2)=0
Answer.
π3,7π6,5π3,11π6
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28.
(tan⁡θ−3)(cot⁡θ+1)=0
Answer.
π3,3π4,4π3,7π4
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30.
2cos2⁡θ−3cos⁡θ=0
Answer.
π6,π2,3π2,11π6
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34.
3tan3⁡θ−tan⁡θ=0
Answer.
0,π6,5π6,π,7π6,11π6
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36.
csc5⁡θ−4csc⁡θ=0
Answer.
π4,3π4,5π4,7π4
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37.
2sin2⁡θ−3sin⁡θ−2=0
Answer.
7π6,11π6
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