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? 

Consider
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
Then
$$
f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}
$$
Therefore
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & |x|<1 \\[6px]
-\dfrac{2}{1+x^2} & |x|>1
\end{cases}
$$
which implies
$$
f(x)=\begin{cases}
c_--2\arctan x & x<-1 \\[6px]
c_0+2\arctan x & -1\le x\le 1 \\[6px]
c_+-2\arctan x & x>1
\end{cases}
$$
and it's easy to see that $c_0=0$, $c_-=-\pi$ and $c_+=\pi$.
Thus the equation can be split into three cases:
$$
\begin{cases}
-2\pi-4\arctan x-\pi x^3=0 \\[4px]
x<-1
\end{cases}
\qquad
\begin{cases}
4\arctan x-\pi x^3=0 \\[4px]
-1\le x\le 1
\end{cases}
\qquad
\begin{cases}
2\pi-4\arctan x-\pi x^3=0 \\[4px]
x>1
\end{cases}
$$
The graph suggests that the solutions are only $-1$, $0$ and $1$ (which are indeed solutions).

Let's consider $\alpha(x)=-2\pi-4\arctan x-\pi x^3$ for $x\le-1$; then
$$
\alpha'(x)=-\frac{4}{1+x^2}-3\pi x^2=-\frac{4+3\pi x^2+3\pi x^4}{1+x^2}
$$
Note that the discriminant of $3\pi t^2+3\pi t+4$ is $9\pi^2-48\pi=3\pi(3\pi-16)<0$ so $\alpha$ is decreasing. As $\alpha(-1)=0$, there are no solutions of your equation for $x<-1$. Similarly, there are no solutions for $x>1$.
In the case $-1\le x\le 1$, the function $\beta(x)=4\arctan x-\pi x^3$ is odd, so we can study it over $[0,1]$. We have
$$
\beta'(x)=\frac{4}{1+x^2}-3\pi x^2=-\frac{3\pi x^4+3\pi x^2-4}{1+x^2}
$$
The derivative vanishes only for
$$
x=\sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}\approx 0.5668
$$
Thus $0$ and $1$ are the only solutions on the interval $[0,1]$; by symmetry,  $0$ and $-1$ are the only solutions on the interval $[-1,0]$.
It's not needed to look for an approximation: the following set of inequalities are equivalent to each other:
\begin{gather}
\sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}<1 \\[6px]
\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}<1 \\[6px]
-3\pi+\sqrt{9\pi^2+48\pi}<6\pi \\[6px]
\sqrt{9\pi^2+48\pi}<9\pi \\[6px]
9\pi^2+48\pi<81\pi^2 \\[6px]
48<72\pi \\[6px]
2<3\pi
\end{gather}
and the last one is true.
