Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$ The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is? 


Let $x= \tan \theta$

$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$

I had to delete the rest of my attempt because it was totally wrong. What are the methods to solve this problem?
As J.G. noted in comments, there are no solutions for $|x|\gt1$. This is because the range of the arcsine function is $[-\pi/2,\pi/2]$, so the left hand side, $2\arcsin\left(2x\over1+x^2\right)$, is never greater than $\pi$ in absolute value, while $\pi|x^3|\gt\pi$ if $|x|\gt1$. We'll come back to this in a moment.

Now after substituting $x=\tan\theta$, the left hand side becomes

$$2\arcsin\left(2\tan\theta\over1+\tan^2\theta\right)=2\arcsin\left(2\sin\theta\cos\theta \right)=2\arcsin(\sin2\theta)$$

It's tempting to "cancel" the the inverse trig function, leaving just $4\theta$.  But that's only correct for $|\theta|\le{\pi\over4}$.  However, that is exactly the domain we need in order to have $|x|\le1$, since $-1\le\tan\theta\le1$ for $-\pi/4\le\theta\le\pi/4$. So we can go ahead and write the equation as

$$4\theta=\pi\tan^3\theta$$

with the understanding that we only care about solutions with $-\pi/4\le\theta\le\pi/4$.

Three solutions jump off the page: $\theta=0$, $\theta=\pi/4$, and $\theta=-\pi/4$. The question is. are there any others?  For this it helps to look at the function $f(\theta)=4\theta-\pi\tan^3\theta$, for which 

$$f'(\theta)=4-3\pi\tan^2\theta\sec^2\theta$$ 

and 
$$f''(\theta)=-3\pi(2\tan\theta\sec^4\theta+2\tan^3\theta\sec^2\theta)=-6\pi\tan\theta\sec^2\theta(2\sec^2\theta-1)$$

It's clear that $f''(\theta)\lt0$ for $0\lt\theta\lt\pi/4$, and easy to see that $f'(0)=4\gt0$ while $f'(\pi/4)=4-6\pi\lt0$. Consequently there are no other solutions in $(0,\pi/4)$, hence, by symmetry, in $(-\pi/4,0)$ either.

To sum things up, $x=0$, $x=1$ and $x=-1$ are the only three solutions to

$$2\arcsin\left(2x\over1+x^2\right)-\pi x^3=0$$
Shouldn't the third line be
$ 2\theta = \dfrac{\pi}{2}\tan^3\theta
$?

The arcsin cancels out the sin.

Wolfy says that
the roots in
$[-\pi/2, \pi/2]$
are
$0$
and
$± 0.785398163397448...$.