## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 42

#### Answer

$x = \frac{\sqrt{5}}{5}$

#### Work Step by Step

$arcsin~2x+arcsin~x = \frac{\pi}{2}$ We can use the following identity: $arcsin~a+arcsin~b = arcsin(a\sqrt{1-b^2}+b\sqrt{1-a^2})$ We can find the value of $x$: $arcsin~2x+arcsin~x = \frac{\pi}{2}$ $arcsin~(2x\sqrt{1-x^2}+x\sqrt{1-(2x)^2}) = \frac{\pi}{2}$ $arcsin~(2x\sqrt{1-x^2}+x\sqrt{1-4x^2}) = \frac{\pi}{2}$ $2x\sqrt{1-x^2}+x\sqrt{1-4x^2} = 1$ $2x\sqrt{1-x^2} = 1 - x\sqrt{1-4x^2}$ $4x^2~(1-x^2) = 1 - 2x\sqrt{1-4x^2}+x^2(1-4x^2)$ $4x^2-4x^4 = 1 - 2x\sqrt{1-4x^2}+x^2-4x^4$ $3x^2 = 1 - 2x\sqrt{1-4x^2}$ $3x^2 - 1 = - 2x\sqrt{1-4x^2}$ $9x^4 - 6x^2+1 = 4x^2(1-4x^2)$ $9x^4 - 6x^2+1 = 4x^2-16x^4$ $25x^4 - 10x^2+1 = 0$ $(5x^2-1)^2 = 0$ $5x^2-1 = 0$ $x^2 = \frac{1}{5}$ $x = \pm \frac{1}{\sqrt{5}}$ $x = \pm \frac{\sqrt{5}}{5}$ We can test both of these candidate solutions in the original equation: $arcsin~[2(- \frac{\sqrt{5}}{5})]+arcsin~(- \frac{\sqrt{5}}{5}) = -\frac{\pi}{2}$ $arcsin~[2(\frac{\sqrt{5}}{5})]+arcsin~(\frac{\sqrt{5}}{5}) = \frac{\pi}{2}$ Therefore: $x = \frac{\sqrt{5}}{5}$

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