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Trigonometry Through Wayfinding and Navigation Across the Pacific

Kamuela E Yong

Section 3.6 Trigonometric Equations - Advanced Techniques

In this section, we will solve trigonometric equations using various trigonometric identities, equations involving multiple angles, and graphical methods.

Subsection 3.6.1 Solving a Trigonometric Equation by Using Fundamental Identities

Example 3.6.1.

Solve \(2\sin^2x+3\cos x-3=0\text{.}\)

Solution.

To solve the equation \(2\sin^2x + 3\cos x - 3 = 0\text{,}\) we want to first write it in terms of only cosine or only sine. By the Pythagorean Identity, we substitute \(\sin^2x=1-\cos^2x\text{,}\) resulting in:
\begin{align*} 2\sin^2x+3\cos x-3 \amp =0 \\ 2(1-\cos^2x)+3\cos x-3 \amp =0 \\ 2-2\cos^2x+3\cos x-3 \amp =0 \\ -2\cos^2x+3\cos x-1 \amp =0 \\ 2\cos^2x-3\cos x+1 \amp =0 \\ (2\cos x-1)(\cos x-1) \amp =0 \text{.} \end{align*}
Thus \(\cos x=\frac{1}{2}\) or \(\cos x=1\text{.}\)
Solving for \(x\) in the first equation gives \(x=\frac{\pi}{3}\) or \(\frac{5\pi}{3}\text{,}\) while the second equations gives \(x=0\) for one period. Finding all solutions, we arrive at \(x=\frac{\pi}{3}+2k\pi\text{,}\) \(\frac{5\pi}{3}+2k\pi\text{,}\) or \(x=0+2k\pi=2k\pi\) for any integer \(k\text{.}\)

Remark 3.6.2.

If factoring trigonometric functions is difficult, we can try substituting a simpler term and then factor. For example, consider the expression \(2\cos^2x-3\cos x+1\) in the previous example. If factoring this expression directly is not clear, we can use the substitution \(A=\cos x\text{.}\) This transforms the expression into \(2A^2-3A+1\text{,}\) which may be easier to factor. Once factored, we can substitute \(\cos x\) back in for \(A\) to obtain the final solution.

Example 3.6.3.

Solve the equation \(\sin(2\theta)+\cos\theta=0\) on the interval \(0\leq\theta\lt2\pi\text{.}\)

Solution.

Notice the first term has \(2\theta\) so we will begin by using the double-angle formula:
\begin{align*} \sin(2\theta)+\cos\theta \amp = 0 \\ 2\sin\theta\cos\theta+\cos\theta \amp = 0 \text{.} \end{align*}
Then we factor \(\cos\theta\) from the terms:
\begin{equation*} \cos\theta(2\sin\theta+1) = 0 \text{.} \end{equation*}
Thus we get
\begin{equation*} \cos\theta=0\quad\mbox{or}\quad2\sin\theta+1=0\text{.} \end{equation*}
When \(\cos\theta=0\) we get \(\theta=\frac{\pi}{2}, \frac{3\pi}{2}\text{.}\)
When \(2\sin\theta+1=0\text{,}\) \(\sin\theta=-\frac{1}{2}\text{,}\) thus \(\theta=\frac{7\pi}{6}, \frac{11\pi}{6}\text{.}\)
Combining these, the solutions are
\begin{equation*} \theta=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\text{.} \end{equation*}

Subsection 3.6.2 Solving a Trigonometric Equation with Multiples of an Angle

Example 3.6.4.

Solve the equation \(2\cos(2\theta)-\sqrt{3}=0\) on the interval \(0\leq\theta\lt2\pi\text{.}\)

Solution.

First we will isolate \(\cos(2\theta)\text{.}\) We begin by rearranging the equation:
\begin{align*} 2\cos(2\theta)-\sqrt{3} \amp = 0 \\ 2\cos(2\theta) \amp = \sqrt{3} \\ \cos(2\theta) \amp = \frac{\sqrt{3}}{2} \text{.} \end{align*}
Since \(\cos(2\theta)\) equals \(\frac{\sqrt{3}}{2}\) for angles \(\frac{\pi}{6}\) and \(\frac{11\pi}{6}\text{,}\) the general solutions for \(2\theta\) are
\begin{equation*} 2\theta=\frac{\pi}{6}+2k\pi\quad\mbox{or}\quad2\theta=\frac{11\pi}{6}+2k\pi \end{equation*}
for any integer \(k\text{.}\)
Note that we have only solved for \(2\theta\text{.}\) Dividing both sides by 2 to find \(\theta\text{,}\) we obtain:
\begin{equation*} \theta=\frac{\pi}{12}+k\pi\quad\mbox{or}\quad\theta=\frac{11\pi}{12}+k\pi\text{.} \end{equation*}
The restriction \(0\leq\theta\lt2\pi\) gives us the following solutions:
\begin{equation*} \frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}\text{.} \end{equation*}

Remark 3.6.5.

In this example, our angle was a double-angle: \(2\theta\text{.}\) When dealing with multiple-angle trigonometric functions, such as \(\cos(2\theta)\text{,}\) it’s essential to understand their graphical behavior. According to Definition 2.1.29, the graph of \(\cos(2\theta)\) undergoes a horizontal compression by a factor of 2, and its period is \(\frac{2\pi}{2}=\pi\text{.}\) Since we are asked to find solutions on the interval \(0\leq\theta\lt2\pi\text{,}\) we will need to consider two periods. Thus, we have four solutions (2 solutions for each period):
\begin{equation*} \frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}\text{.} \end{equation*}
In general, if the trigonometric function has an angle \(k\theta\text{,}\) for some number \(k\text{,}\) we will need to consider the effect of this multiple angle on the period and number of solutions, ensuring that we adjust our solutions accordingly to cover all possible solutions within the given interval.

Example 3.6.6.

Solve the equation \(3\tan\frac{\theta}{2}-\sqrt{3}=0\) on the interval \(0\leq\theta\lt2\pi\text{.}\)

Solution.

We begin by isolating \(\tan\frac{\theta}{2}\text{:}\)
\begin{align*} 3\tan\frac{\theta}{2}-\sqrt{3} \amp = 0 \\ 3\tan\frac{\theta}{2} \amp = \sqrt{3} \\ \tan\frac{\theta}{2} \amp = \frac{\sqrt{3}}{3} \text{.} \end{align*}
Since \(\tan\frac{\theta}{2}\) equals \(\frac{\sqrt{3}}{3}\) for angles \(\frac{\pi}{6}\) and \(\frac{7\pi}{6}\text{,}\) the general solutions for \(\frac{\theta}{2}\) are
\begin{equation*} \frac{\theta}{2}=\frac{\pi}{6}+k\pi\quad\mbox{and}\quad\frac{\theta}{2}=\frac{7\pi}{6}+k\pi \end{equation*}
for any integer \(k\text{.}\)
Solving for \(\theta\text{,}\) we multiply both sides by 2 to obtain:
\begin{equation*} \theta=\frac{\pi}{3}+2k\pi\quad\mbox{and}\quad\theta=\frac{7\pi}{3}+2k\pi\text{.} \end{equation*}
Considering our restriction \(0\leq\theta\lt2\pi\text{,}\) we have only one solution:
\begin{equation*} \theta=\frac{\pi}{3}\text{.} \end{equation*}

Remark 3.6.7.

In this example, the angle was \(\frac{\theta}{2}\text{,}\) which stretches the graph of the tangent function. When we have an angle of the form \(\frac{\theta}{2}\text{,}\) it stretches the period of the tangent function by a factor of 2. Thus, for the given interval, we have only half a period to consider instead of the full period.

Subsection 3.6.3 Solving a Trigonometric Equation with a Graphing Utility

Sometimes we will encounter equations where an exact solution is not possible. However, we may be able to get an approximation to the solution by graphing the equation.

Example 3.6.8.

Use a graphing utility to find the solutions to the equation \(\sin x+\cos x=\frac{1}{2}x\text{.}\) Express the answers in radians, rounded to two decimals.

Solution.

To find the solution to \(\sin x + \cos x = \frac{1}{2}x\text{,}\) we graph the left-hand side and the right-hand side of the equation and identify their intersections. Let \(y_1\) represent the curve for the left-hand side and \(y_2\) represent the curve for the right-hand side:
\begin{equation*} y_1=\sin x+\cos x, \quad y_2=\frac{1}{2}x\text{.} \end{equation*}
Use a graphing utility to plot \(y_1\) and \(y_2\text{.}\)
Figure 3.6.9. Plotting \(y_1=\sin x+\cos x\) and \(\quad y_2=\frac{1}{2}x\text{,}\) corresponding to the left-hand and right-hand sides of the equation, respectively. If you are viewing the PDF or a printed copy, you can scan the QR code or follow the “Standalone” link to explore the interactive version online.
Next, we may need to zoom in or out to better visualize the behavior of the curves. To find their intersection points, calculators often have a TRACE or INTERSECT button or command. In Desmos Graphing Calculator
 1 
Desmos Graphing Calculator
we can click on either curve, and the points of intersection will be highlighted. Hovering the cursor over the intersection will display the coordinates of that point.
The equation \(\sin x + \cos x = \frac{1}{2}x\) has three solutions, which correspond to the points of intersection between the curves \(y_1 = \sin x + \cos x\) and \(y_2 = \frac{1}{2}x\text{.}\) The \(x\)-values of these intersections are
\begin{equation*} x=-2.68, -1.24, 1.71\text{.} \end{equation*}

Exercises 3.6.4 Exercises

Exercise Group.

Solve each equation on the interval \(0\leq\theta\lt2\pi\text{.}\)
1.
\(\sin^2\theta-\cos^2\theta=0\)
Answer.
\(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\)
2.
\(\cos^2\theta-\sin^2\theta=1+\sin\theta\)
Answer.
\(0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\)
3.
\(3\cot^2\theta-4\csc\theta=1\)
Answer.
\(\frac{\pi}{6}, \frac{5\pi}{6}\)
4.
\(\sin^2\theta=5\cos\theta+5\)
Answer.
\(\pi\)
5.
\(\cos^2\theta=3-3\sin\theta\)
Answer.
\(\frac{\pi}{2}\)
6.
\(2\sin^2\theta=3\cos\theta+3\)
Answer.
\(\frac{2\pi}{3}, \pi, \frac{4\pi}{3}\)

Trigonometric Equations Involving Multiples of an Angle.

Solve the given trigonometric equation on the interval \(0\leq\theta\lt2\pi\text{.}\)
7.
\(\cot2\theta=-\sqrt{3}\)
Answer.
\(\frac{5\pi}{12}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{23\pi}{12}\)
8.
\(\sin4\theta=\frac{\sqrt{3}}{2}\)
Answer.
\(\frac{\pi}{12}, \frac{\pi}{6}, \frac{7\pi}{12}, \frac{2\pi}{3}, \frac{13\pi}{12}, \frac{7\pi}{6}, \frac{19\pi}{12}, \frac{5\pi}{3}\)
9.
\(\sqrt{2}\cos2\theta=1\)
Answer.
\(\frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}\)
10.
\(\csc3\theta=2\)
Answer.
\(\frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}\)
11.
\(\sec\frac{3\theta}{2}=-\sqrt{2}\)
Answer.
\(\frac{\pi}{2}, \frac{5\pi}{6}, \frac{11\pi}{6}\)
12.
\(\cos\frac{\theta}{2}-1=0\)
Answer.
\(0\)

Trigonometric Equations Involving Addition or Subtraction Formula.

Use the Addition and Subtraction Formulas to solve each equation on the interval \(0\leq\theta\lt2\pi\text{.}\)
13.
\(\sin\theta\cos2\theta+\cos\theta\sin2\theta=\frac{\sqrt{3}}{2}\)
Answer.
\(\frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}, \frac{13\pi}{9}, \frac{14\pi}{9}\)
14.
\(\sin3\theta\cos2\theta-\cos3\theta\sin2\theta=-\frac{1}{2}\)
Answer.
\(\frac{7\pi}{6}, \frac{11\pi}{6}\)
15.
\(\cos3\theta\cos\theta+\sin3\theta\sin\theta=-\frac{\sqrt{2}}{2}\)
Answer.
\(\frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}\)
16.
\(\cos\theta\cos3\theta-\sin\theta\sin3\theta=1\)
Answer.
\(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\)

Trigonometric Equations Involving Double-Angle or Half-Angle Formula.

Use the Double-Angle and Half-Angle Formulas to solve each equation on the interval \(0\leq\theta\lt2\pi\text{.}\)
17.
\(\sin2\theta=\cos\theta\)
Answer.
\(\frac{\pi}{6}, \frac{\pi}{2},\frac{5\pi}{6}, \frac{3\pi}{2}\)
18.
\(\cos2\theta=\cos\theta\)
Answer.
\(0, \frac{2\pi}{3}, \frac{4\pi}{3}\)
19.
\(\cos2\theta+\cos\theta-2=0\)
Answer.
\(0\)
20.
\(\tan2\theta=-2\sin\theta\)
Answer.
\(0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}\)
21.
\(\cos2\theta+2=2\sin^2\theta\)
Answer.
\(\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\)
22.
\(\tan\frac{\theta}{2}=\sin\theta\)
Answer.
\(0, \frac{\pi}{2}, \frac{3\pi}{2}\)

Trigonometric Equations Involving Sum-to-Product Formula.

Use the Sum-to-Product Formulas to solve each equation on the interval \(0\leq\theta\lt2\pi\text{.}\)
23.
\(\sin3\theta+\sin\theta=0\)
Answer.
\(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\)
24.
\(\sin6\theta-\sin2\theta=\cos4\theta\)
Answer.
\(\frac{\pi}{12}, \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{12}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{13\pi}{12}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{17\pi}{12}, \frac{13\pi}{8}, \frac{15\pi}{8}\)
25.
\(\cos4\theta-\cos2\theta=0\)
Solution.
\(0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}\)
26.
\(\cos3\theta+\cos\theta=0\)
Solution.
\(\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}\)

Exercise Group.

Use a graphing utility to solve each equation. Express your solutions in radians, rounded to two decimals.
27.
\(\sin(2x)=4\cos x+x\)
Answer.
\(x=-1.34, 2.63, 3.90\)
28.
\(\sin x-x=\cos x\)
Answer.
\(x=-1.26\)
29.
\(x^2=\cos x\)
Answer.
\(x=-0.82, 0.82\)
30.
\(x^3+2x^2=\cos(2x)\)
Answer.
\(x=-2.11, -0.56, 0.48\)
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