Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 35


$x = \{\frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$

Work Step by Step

$\sqrt{2}~sin~3x-1 = 0$ $sin~3x = \frac{1}{\sqrt{2}}$ $sin~3x = \frac{\sqrt{2}}{2}$ $3x = arcsin(\frac{\sqrt{2}}{2})$ $3x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{17\pi}{4}, \frac{19\pi}{4}$ $x = \frac{\pi}{12}, \frac{3\pi}{12}, \frac{9\pi}{12}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$ $x = \frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$ For the domain $[0, 2\pi]$, the solution set is as follows: $x = \{\frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$
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.