Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 41



Work Step by Step

$\sec^{-1}{\frac{2\sqrt3}{3}}=\frac{\pi}{6}$, because $\sec{\frac{\pi}{6}}=\frac{2\sqrt3}{3}$ and $\frac{\pi}{6}$ is in the range of $\sec^{-1}{x}$, which is $[0,\pi]$ \ $\left\{\frac{\pi}{2}\right\}.$
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