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: 24



Work Step by Step

$cos(2x)=\frac{-1}{2}$ $2x=cos^{-1}(\frac{-1}{2})$ We know $cos(x)$ is Negative in quardent $II$ and quardent $III$ The period of the cosine function is $360^o$ $2x=180^o-60^o\;\;\;\;\;\;\;or\;\;\;\;2x=360^o+120^o\;\;\;or\;\;\;\;\;\;\;2\theta=180^o+60^o\;\;\;\;\;\;\;or\;\;\;\;\;\;2x=360^o+240^o$ $2x=120^o\;\;\;\;\;\;\;or\;\;\;\;2x=480^o\;\;\;or\;\;\;\;\;\;\;2x=240^o\;\;\;\;\;\;\;or\;\;\;\;\;\;2x=600^o $ $x=60^o\;\;\;\;\;\;\;or\;\;\;\;x=120^o\;\;\;\;\;or\;\;\;\;\;\;\;x=240^o\;\;\;\;\;\;\;or\;\;\;\;\;\;\;x=300^o $ $x=\{60^o\;,120^o,240^o,300^o\}\;\;\;\;\;\;\;\;\;\;$
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.