Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Chapter Test - Page 508: 26

Answer

$0, 1.911, \pi, 4.373$

Work Step by Step

Step 1. $-3cos(\frac{\pi}{2}-\theta)=tan(\theta) \longrightarrow -3sin(\theta)=sin(\theta)/cos(\theta)$ Thus we have $sin(\theta)=0$ or $cos(\theta)=-\frac{1}{3}$ Step 2. For $sin(\theta)=0$, we have $\theta=k\pi$ Step 3. For $cos(\theta)=-\frac{1}{3}$, we have $\theta=2k\pi+cos^{-1}(-\frac{1}{3})\approx2k\pi+1.911$ and $\theta=2k\pi+2\pi-cos^{-1}(-\frac{1}{3})\approx2k\pi+4.373$ Step 4. Within $[0,2\pi)$, we have $\theta=0, 1.911, \pi, 4.373$
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.