Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 73



Work Step by Step

Step 1. Using the identity $cos(2x)=2cos^2(x)-1$, we have $2cos^2(x)+5cos(x)+2=0$ or $(cos(x)+2)(2cos(x)+1)=0$, which gives $cos(x)=-2$ and $cos(x)=-\frac{1}{2}$ Step 2. For $cos(x)=-2$, there is no solution. Step 3. For $cos(x)=-\frac{1}{2}$ , we can find all x-values in $[0,2\pi)$ as $x=\frac{2\pi}{3},\frac{4\pi}{3}$ Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{2\pi}{3},\frac{4\pi}{3}\}$
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