The number of solution of the equation $2{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-\pi x^3=0$ is equal to

We have $2{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-\pi x^3=0$

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

Now we know that

${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\{\begin{array}{ll}-2{\tan }^{-1}x-\pi ,& x<-1\\ 2{\tan }^{-1}x,& -1\le x\le 1\\ -2{\tan }^{-1}x+\pi ,& x>1\end{array}$

To find the number of solutions of the equation, we draw the graphs of 

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

Clearly, there are 3 solutions of the equation $\{-1,0,1\}$