Q:

Solve $\arcsin (\frac{2x}{x^2+1})= \arctan(x)$

I am trying to solve the next equation:
$$
\arcsin\left(\frac{2x}{x^2+1}\right)=\arctan(x)
$$
When I pose $\tan \left(\frac{t}{2}\right) =x$
I obtain, after some simplifications, $t = \frac{t}{2}$.
I know $\frac{\pi}{3},\frac{-\pi}{3}$ are answers but can't seem to get them.
Can you help me? 

A:

Since $\sin\arctan{x}=\frac{x}{\sqrt{1+x^2}},$ we need to solve
$$\frac{2x}{1+x^2}=\frac{x}{\sqrt{1+x^2}},$$ which gives the answer:
$$\{0,\sqrt3,-\sqrt3\}.$$

