JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 4)
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 :
0
1
$$\frac{1}{2}$$
$$-\frac{1}{2}$$
Explanation
$${\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x$$
For Domain : $$x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]$$
$${\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)$$
$$ \Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x$$
$$ \Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)$$
$$ \Rightarrow 4{x^3} = x$$
$$ \Rightarrow x = 3,\, \pm \,{1 \over 2}$$
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