Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises - Page 721: 38



Work Step by Step

Recognize the RHS of a double angle identity for cosine, $\cos 2\theta=0.$ Since $0^{o} \leq \theta < 360^{o},$ then $0^{o} \leq 2\theta < 720^{o},$ So, $ 2\theta$ can be $90^{o}, 270^{o}, $ $90^{o}+360^{o}=450^{o}$ or $270^{o}+360^{o}=630^{o}.$ Halving these angles, $\theta=45,135^{o},225^{o},315^{o}$ Solution set = $\{45,135^{o},225^{o},315^{o}\}$ ------- Another method: recognize a difference of squares $(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=0$ .. and apply the zero factor principle ... 1. $\cos\theta=-\sin\theta\quad $or 2. $\cos\theta=\sin\theta$, yielding the same solution set.
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.