#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 3 Subquestion (i) Maths Textbook Solution.

Answer: $x= +\frac{1}{2}\sqrt{\frac{3}{7}}$
Given: $\sin^{-1}x+\sin^{-1}2x= \frac{\pi }{3}$
Hint: Using the formula   $\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
Solution:  We know that  $\sin^{-1}\left ( \frac{\sqrt{3}}2{} \right )= \frac{\pi }{3}$
$\Rightarrow \sin^{-1}2x=\sin^{-1}\frac{\sqrt{3}}{2}-\sin^{-1}x$
Using the formula, $\sin^{-1}x-\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ]$
$\Rightarrow \sin^{-1}2x= \sin^{-1}\left [ \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-x\sqrt{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}} \right ]$
$\Rightarrow 2x=\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}-\frac{x}{2}$
$\Rightarrow 2x+\frac{x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow \frac{5x}{2}= \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}$
$\Rightarrow 5x= \sqrt{3}\sqrt{1-x^{2}}$
Squaring on both sides, we get
$\Rightarrow 25x^{2}= 3\left ( 1-x^{2} \right )$
$\Rightarrow 25x^{2}= 3-3x^{2}$
$\Rightarrow 28x^{2}= 3$$x= \pm \frac{1}{2}\sqrt{\frac{3}{7}}$
As $x= - \frac{1}{2}\sqrt{\frac{3}{7}}$is not satisfying the equation
Hence  $x= +\frac{1}{2}\sqrt{\frac{3}{7}}$