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 - Review Exercises - Page 507: 70

Answer

$0,\pi,\frac{2\pi}{3},\frac{4\pi}{3}$

Work Step by Step

Step 1. $sin(\theta)+2sin(\theta)cos(\theta)=0 \longrightarrow sin(\theta)(1+2cos(\theta))=0 \longrightarrow sin(\theta)=0$ and $cos(\theta)=-\frac{1}{2}$ Step 2. For $sin(\theta)=0$, we have $\theta=2k\pi$ and $\theta=2k\pi+\pi$ Step 3. For $cos(\theta)=-\frac{1}{2}$, we have $\theta=2k\pi+\frac{2\pi}{3}$ and $\theta=2k\pi+\frac{4\pi}{3}$ Step 4. Within $[0,2\pi)$, we have $\theta=0,\pi,\frac{2\pi}{3},\frac{4\pi}{3}$
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