Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 340: 32


$$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$

Work Step by Step

Given $$ \cos 2x \cos x - \sin 2x \sin x=\frac{ \sqrt{2}}{2},\ \ 0\leq x<2\pi $$ Then using $\cos(A+B)= \cos A \cos B - \sin A \sin B$ , we get \begin{align*} \cos 2x \cos x - \sin 2x \sin x&=\frac{ \sqrt{2}}{2}\\ \cos(3x)&=\frac{ \sqrt{2}}{2}\\ \end{align*} First we find all possible solutions for $x$ $$ 3x=\frac{\pi}{4}+2k\pi \ \Rightarrow x= \frac{\pi}{12}+ \frac{2k\pi}{3}$$ To find those solutions that lie in the interval $0\leq x<2\pi$ we let k take on values of 0, 1, and 2; we get $$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.