Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.3 Trigonometric Equations - 7.3 Assess Your Understanding - Page 467: 66

Answer

$\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$

Work Step by Step

Step 1. Rewrite the equation $1-sin^2\theta-sin^2\theta+sin\theta=0 \longrightarrow 2sin^2\theta-sin\theta-1=0 \longrightarrow (sin\theta-1)(2sin\theta+1)=0$. Thus $sin\theta=1$ or $sin\theta=-\frac{1}{2}$ Step 2. For $sin\theta=1$, we have $\theta=\frac{\pi}{2}$ Step 3. For $sin\theta=-\frac{1}{2}$, we have $\theta=\frac{7\pi}{6},\frac{11\pi}{6}$
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