Answer
$cos^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{6}$
Work Step by Step
The range of $cos^{-1}~x$ is $[0,\pi]$. So,
$cos^{-1}~(\frac{\sqrt 3}{2})=\frac{\pi}{6}$, because $cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$ and $0\leq\frac{\pi}{6}\leq\pi$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 6 - Review Questions - Page 501: 96
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