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: 66


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

Work Step by Step

Step 1. Using the identity $sin^2x=1-cos^2x$, we have $4-4cos^2x+4cos(x)-5=0$ or $4cos^2x-4cos(x)+1=0$, or $(2cos(x)-1)^2=0$, which gives solutions as $cos(x)=\frac{1}{2}$ (multiplicity 2) Step 2. For $cos(x)=\frac{1}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{3},\frac{5\pi}{3}$ Step 3. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{3},\frac{5\pi}{3} \}$
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