Calculus 8th Edition

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

Appendix D - Trigonometry - D Exercises - Page A33: 71


$0, \pi, 2 \pi$

Work Step by Step

Need to find the range for $x$ for the equation $sinx=tanx$ $sinx-tanx=0$ $sinx-\frac{sinx}{cosx}=0$ $sinx(1-\frac{1}{cosx})=0$ Here, $sinx=0$ gives $x =0, \pi, 2\pi$ and $(1-\frac{1}{cosx})=0$ gives $cosx=1$ $x =0, 2\pi$ Hence, $0, \pi, 2 \pi$
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