Answer
$\frac{\pi}{6}$
Work Step by Step
Letting $sec^{-1}(\frac{2}{\sqrt 3})=t, 0\lt t \lt \pi$, we have $sec(t)=\frac{2}{\sqrt 3}$
Thus $cos(t)=\frac{\sqrt 3}{2}$ and $t=\frac{\pi}{6}$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 7 - Analytic Trigonometry - Chapter Review - Chapter Test - Page 507: 1
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