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#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 3  Maths Textbook Solution.

$x= \frac{1}{2}$
Given:
$\cos^{-1}\sqrt{3x}+\cos^{-1}x= \frac{x}{2}$
Hint:
We are applying the formula,
$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right ) \right ]$
Solution:
We have
$\cos^{-1}\sqrt{3x}+\cos^{-1}x= \frac{x}{2}$
Using formula written on the hint, we get
$\Rightarrow \cos^{-1}\left [ \sqrt{3x}\times x-\left ( \sqrt{1-\left ( \sqrt{3x} \right )^{2}} \right )\left ( \sqrt{1-x^{2}} \right ) \right ]= \frac{x}{2}$
$\Rightarrow \cos^{-1}\left [ \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right ) \right ]= \frac{x}{2}$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )= \cos \frac{x}{2}$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )= 0\; \; \; \; \; \; \left [ \because \cos \frac{x}{2}= 0 \right ]$
$\Rightarrow \sqrt{3x^{2}}-\left ( \sqrt{1-3x ^{2}} \right )\left ( \sqrt{1-x^{2}} \right )$
Squaring on both sides, we get
$\Rightarrow 3x^{4}-\left ( 1-3x ^{2} \right )\left ( 1-x^{2} \right )$
$\Rightarrow 3x^{4}= 1-x ^{2} -3x^{2}+3x^{4}$
$\Rightarrow 1-4x^{2}= 0$
$\Rightarrow 4x^{2}= 1$
$\Rightarrow x^{2}= \frac{1}{4}$
$\Rightarrow x=\pm \frac{1}{2}\; \; \; \; \; \; \; \; \; [ As -\frac{1}{2} is\: not\: satisf\! ying \: the\: equation]$
Hence the result is $x= \frac{1}{2}$