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: 73

Answer

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

Work Step by Step

Step 1. $4(1-cos^2(\theta))=1+4cos(\theta) \longrightarrow 4cos^2(\theta))+4cos(\theta)-3=0 \longrightarrow (2cos(\theta)+3)(2cos(\theta)-1)=0 $. Thus $cos(\theta)=\frac{1}{2}$ or $cos(\theta)=-\frac{3}{2}$ Step 2. For $cos(\theta)=\frac{1}{2}$, we have $\theta=2k\pi+\frac{\pi}{3}$ and $\theta=2k\pi+\frac{5\pi}{3}$ Step 3. For $cos(\theta)=-\frac{3}{2}$, there is no real solution. Step 4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{3}, \frac{5\pi}{3}$
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