Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Cumulative Review Exercises - Page 1038: 24


$0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

Work Step by Step

Step 1. Since $sin2\theta = 2 sin\theta cos\theta$, we have $ 2 sin\theta cos\theta - sin\theta =0$ Step 2. Factor the equation to get $sin\theta (2cos\theta -1)=0$; thus $sin\theta=0$ and $cos\theta=\frac{1}{2}$ Step 3. For $sin\theta=0$, we have solutions in $[0,2\pi)$ as $\theta = 0. \pi$ Step 4. For $cos\theta=\frac{1}{2}$, we have solutions in $[0,2\pi)$ as $\theta =\frac{\pi}{3}, \frac{5\pi}{3}$ Step 5. The solutions in $[0,2\pi)$ are $\theta =0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$
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