Solving Trigonometric Equations

Section 7.6 Solving Trigonometric Equations

In this section, we will learn to solve trigonometric equations by hand and with a calculator.

Subsection Textbook Reference

This relates to content in §9.5 of Algebra and Trigonometry 2e.

Exercises Preparation Exercises

Answer the following without using a calculator. These questions are intended to check the prerequisite skills needed to complete the rest of the lab.

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1.

Solve the equation $2x^2 + 5x - 6=0$ by hand.

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2.

Carlos and Reba were having a discussion about the equation $\sin(\theta)=\frac{1}{2}\text{.}$ The answers below are either "both Reba and Carlos are right", "only Reba is right", "only Carlos is right" or "neither Reba nor Carlos are right".

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(a)

Reba said that $\frac{5\pi}{6}$ made the equation true. Carlos said that $\frac{\pi}{6}$ made the equation true. Who was right?

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(b)

Reba said that $-\frac{5\pi}{6}$ made the equation true. Carlos said that $-\frac{\pi}{6}$ made the equation true. Who was right?

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(c)

Reba said that $-\frac{\pi}{6}$ was the same as $\frac{11\pi}{6}\text{.}$ Carlos said that $-\frac{\pi}{6}$ is not the same as $\frac{11\pi}{6}\text{.}$ Who was right?

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(d)

Reba said that $-\frac{7\pi}{6}$ made the equation true. Carlos said that $-\frac{\11pi}{6}$ made the equation true. Who was right?

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(e)

Reba said that the value of $\sin^{-1}\left(\frac{1}{2}\right)$ was $\frac{5\pi}{6}\text{.}$ Carlos said that the value of $\sin^{-1}\left(\frac{1}{2}\right)$ was $\frac{\pi}{6}\text{.}$ Who was right?

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(f)

Reba said that thinking about the equation $\sin(\theta)=\frac{1}{2}$ is the same as thinking about angles that have a $y$-value of $\frac{1}{2}\text{.}$ Carlos said that thinking about the equation $\sin(\theta)=\frac{1}{2}$ is the same as thinking about angles that have an $x$-value of $\frac{1}{2}\text{.}$ Who was right?

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(g)

Reba said there are two solutions to the equation $\sin(\theta)=\frac{1}{2}\text{.}$ Carlos said that there are infinitely many solutions to the equation $\sin(\theta)=\frac{1}{2}\text{.}$ Who was right?

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Exercises Practice Exercises

1.

Practice solving the equations, on the given intervals, with your memorized unit circle values.

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(a)

$\cos(\theta) = \frac{\sqrt{2}}{2}$ on the interval $\left[0,3\pi\right]\text{.}$

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(b)

$\cos(\theta) = \frac{\sqrt{2}}{2}$ on the interval $\left[-\pi,\pi\right]\text{.}$

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(c)

$\cos(\theta) = \frac{\sqrt{2}}{2}$ on the interval $\left(-\infty,\infty\right)\text{.}$

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(d)

$\sin(\theta) = \frac{\sqrt{3}}{2}$ on the interval $\left[\pi,4\pi\right]\text{.}$

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(e)

$\sin(\theta) = \frac{\sqrt{3}}{2}$ on the interval $\left[-\pi,\pi\right]\text{.}$

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(f)

$\sin(\theta) = \frac{\sqrt{3}}{2}$ on the interval $\left(-\infty,\infty\right)\text{.}$

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2.

Solve the trigonometric equations by hand without a calculator.

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(a)

$1 - 2 \sin(\chi) = 0$ on the interval $\left[0,2\pi\right]\text{.}$

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(b)

$2 \cos^2(\tau) + \cos(\tau) = 1$ on the interval $\left[-\pi,\pi\right]\text{.}$

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(c)

$\sqrt{3}\tan(\kappa)+3=0$ on the interval $\left[0,2\pi\right]\text{.}$

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(d)

$4 \cos^2(\alpha) = 1$ on the interval $\left[-2\pi,2\pi\right]\text{.}$

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(e)

$2 \sin(3\psi) = -\sqrt{3}$ on the interval $\left[0,\pi\right]\text{.}$

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(f)

$2\cos(5\omega)+\sqrt{2}=0$ on the interval$\left[0,\pi\right]\text{.}$

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3.

Solve the trigonometric equations by hand, and then use a calculator to approximate the solutions.

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(a)

$1 - 3 \sin(\chi) = 0$ on the interval $\left[0,2\pi\right]\text{.}$

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(b)

$\sin^2(\phi) = 0.6$ on the interval $\left[0,2\pi\right]\text{.}$

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4.

If you were trying to solve the equation $2\sin^2(x) = \sin(x) + 1$ on the interval $[-\pi,pi]\text{,}$ and you had a graph of $y = 2\sin^2(x)$ and $y = \sin(x) + 1\text{,}$ what would you look for on the graph to solve the equation? Create that graph, then write out the solution set based on your work.

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$described in detail following the image$

A blank graph.

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5.

Solve the trigonometric equations using identities by hand.

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(a)

$\sin(\sigma) = -cos(\sigma)$ on the interval $\left(-\infty,\infty\right)\text{.}$

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(b)

$\sin(2x) = cos(2x)$ on the interval $\left[-\pi,\pi\right]\text{.}$

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(c)

$\sin(x) = cos(2x)$ on the interval $\left[-\pi,\pi\right]\text{.}$

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(d)

$\sin(4\theta) - \sin(2\theta) = 0$ on the interval $\left[-\pi,\pi\right]$ by first using the difference-to-product identity $\sin(\alpha) - \sin(\beta) = 2\sin\left(\frac{\alpha - \beta}{2}\right) \cos\left(\frac{\alpha + \beta}{2}\right)\text{.}$

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Exercises Exit Exercises

1.

Explain why you know that the equation $\cos(x) = 3$ couldn’t possibly have any real solutions.

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2.

Solve the equation $2\sin^2(\beta)-\sin(\beta)=1$ on the interval $\left[-\pi,\pi\right]$ without a calculator.

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3.

Solve the equation $\sin^2(\phi) = 1$ on the interval $\left(-\infty,\infty\right)$ without a calculator.

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4.

Explain the solving process, in English, for the equation $\cos(\theta) = \frac{1}{2}$ on the interval $\left[-2\pi,2\pi\right]$ step by step as if you were explaining it to someone in your class who wanted to understand today’s lesson more deeply. Actually solving the equation isn’t necessary.

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Reflection Reflection

1.

On a scale of 1-5, how are you feeling with the concepts related to the graphical behaviors of functions?

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