Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 589: 75

Answer

$\frac{4\pi}{3}, \frac{5\pi}{3}$

Work Step by Step

Given $sin(s)=-\frac{\sqrt 3}{2}$, we can find a reference angle as $s_0=sin^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{3}$. In $[0,2\pi)$, there are two angles $s=\pi+\frac{\pi}{3}, 2\pi-\frac{\pi}{3}$ or $s=\frac{4\pi}{3}, \frac{5\pi}{3}$
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.