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? 
 A: 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
$$2\arcsin\left(2x\over1+x^2\right)-\pi x^3=0$$
A: 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.
A: 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 $x=-1,0,1$.
A: 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
$± 0.785398163397448...$.
