Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 41


$x = -\frac{1}{2}$

Work Step by Step

$arcsin~2x+arccos~x = \frac{\pi}{6}$ $arcsin~2x= \frac{\pi}{6}-arccos~x $ $2x= sin(\frac{\pi}{6}-arccos~x)$ $2x= sin(\frac{\pi}{6})~cos(-arccos~x)+cos(\frac{\pi}{6})~sin(-arccos~x)$ $2x= (\frac{1}{2})~(x)+\frac{\sqrt{3}}{2}~(-\sqrt{1-x^2})$ $\frac{3}{2}x= \frac{\sqrt{3}}{2}~(-\sqrt{1-x^2})$ $\frac{9}{4}x^2= \frac{3}{4}~(1-x^2)$ $3x^2= \frac{3}{4}$ $x^2 = \frac{1}{4}$ $x = \pm\frac{1}{2}$ We can check our two candidate solutions: $arcsin~2(-\frac{1}{2})+arccos~(-\frac{1}{2}) = -\frac{\pi}{2}+\frac{2\pi}{3} = \frac{\pi}{6}$ $arcsin~2(\frac{1}{2})+arccos~(\frac{1}{2}) = \frac{\pi}{2}+\frac{\pi}{3} = \frac{5\pi}{6}$ The solution is: $x = -\frac{1}{2}$
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