Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 36

Answer

$x = \frac{\sqrt{3}}{2}$

Work Step by Step

$sin^{-1}~x+tan^{-1}~\sqrt{3} = \frac{2\pi}{3}$ $sin^{-1}~x+\frac{\pi}{3} = \frac{2\pi}{3}$ $sin^{-1}~x = \frac{\pi}{3}$ $x = sin~ \frac{\pi}{3}$ $x = \frac{\sqrt{3}}{2}$
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