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 339: 9

Answer

$\theta=\{60^o,180^o,300^o\}$

Work Step by Step

$cos(3\theta)=-1$ $3\theta=cos^{-1}(-1)$ We know $cos(\theta)$ is negative in quardent $II$ and quardent $III$ The period of the cosine function is $360^o$ $3\theta=180^o\;\;\;\;or\;\;\;\;\;\;3\theta=360^o+180^o\;\;\;\;or\;\;\;\;\;\;3\theta=720^o+180^o\;\;\;\;\;\;\;\;\;\;$ $\theta=60^o\;\;\;\;or\;\;\;\;\;\;\theta=180\;\;\;\;or\;\;\;\;\;\;\theta=300^o\;\;\;\;\;\;\;\;\;\;\;\;\;$ $\theta=\{60^o,180^o,300^o\}$
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