Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 69

Answer

$\frac{ \pi}{2},\frac{3 \pi}{2}$,$\frac{\pi}{6},\frac{5 \pi}{6}$

Work Step by Step

Need to find the range for $x$ for the equation $sin2x=cosx$ $2sinxcosx-cosx=0$ $cosx(2sinx-1)=0$ Here, $cosx=0$ gives $x =\frac{ \pi}{2},\frac{3 \pi}{2}$ and $2sinx-1=0$ gives $sinx=\frac{1}{2}$ Since, $sin>0$ on first and second quadrant , thus, x must lie on $\frac{\pi}{6},\frac{5 \pi}{6}$ Hence, $\frac{ \pi}{2},\frac{3 \pi}{2}$,$\frac{\pi}{6},\frac{5 \pi}{6}$
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