The no. of solutions of the equation ${\sin }^{-1}\mathrm{x}+{\sin }^{-1}2\mathrm{x}=\frac{\mathrm{\pi }}{3}\text{ is }$

The given equation is ${\sin }^{-1}x+{\sin }^{-1}2\mathrm{x}=\frac{\mathrm{\pi }}{3}$

This can be written as

${\sin }^{-1}x=\frac{\pi }{3}-{\sin }^{-1}\left(2x\right)$

apply sine function both sides

$\begin{array}{rcl}x& =& \sin \left(\frac{\mathrm{\pi }}{3}-{\sin }^{-1}\left(2x\right)\right)\\ & =& \frac{\sqrt{3}}{2}\sqrt{1-4x^2}-\frac{1}{2}\left(2x\right)\\ 2x& =& \frac{\sqrt{3}}{2}\sqrt{1-4x^2}\\ 4x& =& \sqrt{3-12x^2}\end{array}$

Squaring on both sides

$\begin{array}{rcl}16x^2& =& 3-12x^2\\ 28x^2& =& 3\\ x& =& \sqrt{\frac{3}{28}}\end{array}$

Take posiitve value only, if it is negative, the angle must be negative in the given equation

so the number of solutions for the given equaiton is only one