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: 74


$0\leq x\lt\frac{2\pi}{3},\frac{4 \pi}{3}\lt x\leq 2\pi$

Work Step by Step

$2cosx+1>0$ Here, $2cosx+1=0$ gives $cosx=-\frac{1}{2}$ cosx is equal to .5 for x=$\frac{2\pi}{3},\frac{4 \pi}{3}$ Hence, the solution for the given equation is $0\leq x\lt\frac{2\pi}{3},\frac{4 \pi}{3}\lt x\leq 2\pi$
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