Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 709: 62


$0,\frac{\pi}{6},\pi, \frac{11\pi}{6}$.

Work Step by Step

Step 1. Rewrite the equation as $2sin(x)cos(x)=\sqrt 3sin(x)$ or $sin(x)(2cos(x)-\sqrt 3)=0$ Step 2. For $sin(x)=0$, we have solutions in $[0,2\pi)$ as $x=0,\pi$ Step 3. For $cos(x)=\frac{\sqrt 3}{2}$, we have solutions in $[0,2\pi)$ as $x=\frac{\pi}{6}, \frac{11\pi}{6}$ Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=0,\frac{\pi}{6},\pi, \frac{11\pi}{6}$.
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