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: 12



Work Step by Step

$sin(2x)=\frac{\sqrt{2}}{2}$ $2x=sin^{-1}(\frac{\sqrt{2}}{2})$ We know $sin(x)$ is positive in quardent $I$ and quardent $II$ The period of the sine function is $2\pi$ $2x=\frac{\pi}{4}\;\;\;\;or\;\;\;\;\;\;2x=2\pi+\frac{\pi}{4}\;\;\;\;or\;\;\;\;\;\;2x=\pi -\frac{\pi}{4}\;\;\;\;\;\;or\;\;\;\;2x=2\pi + \frac{3\pi}{4} $ $x=\frac{\pi}{8}\;\;\;\;or\;\;\;\;\;\;x=\frac{9\pi}{8}\;\;\;\;or\;\;\;\;\;\;x=\frac{3\pi}{8}\;\;\;\;\;\;\;\;or\;\;\;\;\;x=\frac{11\pi}{8}$ $x=\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\}$
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.