Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$ is equal to :
Solution
<p>${\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x$</p>
<p>For Domain : $x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]$</p>
<p>$${\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)$$</p>
<p>$\Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x$</p>
<p>$\Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)$</p>
<p>$\Rightarrow 4{x^3} = x$</p>
<p>$\Rightarrow x = 3,\, \pm \,{1 \over 2}$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
This question is part of PrepWiser's free JEE Main question bank. 65 more solved questions on Inverse Trigonometric Functions are available — start with the harder ones if your accuracy is >70%.