Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 73

Answer

$0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$

Work Step by Step

$sinx\leq \frac{1}{2}$ For $sinx=\frac{1}{2}$ $sinx=sin\frac{\pi}{6}$ Thus, the complete solutions for $x=\frac{\pi}{6},\frac{5\pi}{6}$ Hence, the given equation has solutions: $0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
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.