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: 57

Answer

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

Work Step by Step

Step 1. Given the equation $tan(x)=2cos(x)tan(x)$ we have $tan(x)(2cos(x)-1)=0)$, Step 2. For $tan(x)=0$, we can find the solutions in $[0,2\pi)$ as $x=0,\pi$ Step 3. For $cos(x)=\frac{1}{2}$, we can find the solutions in $[0,2\pi)$ as $x=\frac{\pi}{3}, \frac{5\pi}{3} $ Step 4. Thus, we can find all the solutions in $[0,2\pi)$ as $0,\frac{\pi}{3}, \pi, \frac{5\pi}{3} $
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