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? 

Let
$$
f(x)=2\arcsin\frac{2x}{1+x^2}-\pi x^3.
$$
Then $f'(x)=-3\pi x^2-\frac{4}{1+x^2}<0$ if $|x|>1$, namely $f(x)$ is decreasing in $(-\infty,-1)$ and $(1,\infty)$. Also
$$ f'(x)=-3\pi x^2+\frac{4}{1+x^2}, |x|<1$$
and
$$ f''(x)=-\frac{2x[4+3\pi(x^2+1)^2]}{(x^2+1)^2}, |x|<1. $$
So $f''(x)>0$ if $x\in(-1,0)$ and $f''(x)<0$ if $(0,1)$ and hence $f(x)$ is strictly concave in $(-1,0)$ and strictly convex in $(0,1)$. Note that $f(-1)=f(0)=f(1)=0$. Thus $f(x)=0$ only has three roots $x=-1,0,1$.
