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: 42

Answer

$\dfrac{2\pi}{3}$

Work Step by Step

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