Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises - Page A 33: 73


$0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$

Work Step by Step

$sinx\leq \frac{1}{2}$ For $sinx=\frac{1}{2}$ $.5=sin\frac{\pi}{6}=sin\frac{5 \pi}{6}$ Thus $x=\frac{\pi}{6},\frac{5\pi}{6}$ Hence, the given equation has solutions: $0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
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