## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 23

#### Answer

The solution set to the given equation is $$\theta=\{66.5^\circ, 112.5^\circ, 247.5^\circ,292.5^\circ\}$$

#### Work Step by Step

$$\sqrt2\cos2\theta=-1$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $2\theta$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $2\theta$, the inequality would be $$0^\circ\le2\theta\lt720^\circ$$ Thus, the corresponding interval for $2\theta$ is $[0^\circ,720^\circ)$. 2) Now we examine the equation: $$\sqrt2\cos2\theta=-1$$ $$\cos2\theta=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$$ Over interval $[0^\circ,720^\circ)$, there are 4 values whose sine equals $-\frac{\sqrt2}{2}$, which are $\{135^\circ, 225^\circ, 495^\circ,585^\circ\}$ Therefore, $$2\theta=\{135^\circ, 225^\circ, 495^\circ,585^\circ\}$$ It follows that $$\theta=\{66.5^\circ, 112.5^\circ, 247.5^\circ,292.5^\circ\}$$ This is the solution set of the equation.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.