How to solve $2 \tan x / (1 - (\tan x)^2) = (\sin 2x)^2$? $$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$
I tried a lot but I get nowhere
 A: Using Double Angle formula for tangent, $$\tan2x=\sin^22x$$
$$\sin2x(1-\sin2x\cos2x)=0$$
Now $\sin2x\cos2x=\dfrac{\sin4x}2\le\dfrac12$
Now $\sin2x=0\implies2x=n\pi $ where $n$ is any integer
A: $\frac{2\tan x}{1-\tan^2x}=(\sin 2x)^2$
$\tan 2x=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}-\sin^2 2x=0$
$\frac{\sin 2x}{\cos 2x}(1-\sin 2x \cos 2x)=0$
either $\sin 2x=0$ or $1-\sin 2x \cos 2x=0$
$\sin 2x=0$ gives $x=\frac{n\pi}{2}$
$1-\sin 2x \cos 2x=0\Rightarrow\sin 4x=2$ which is not possible as $\sin$ cannot be greater than 1.
A: $\sin{2x}=\frac{2\tan x}{1+\tan^2{x}}$
One solution is instantly, $x=n\pi$..
For other solutions, $\tan{x} \ne 0$, you can cross multiply and obtain a quartic equation in $\tan{x}$. I am typing on my mobile, so not working the whole solution out right away. But hope this enough for you to proceed.
A: $$\frac {2\tan {x}}{1-(\tan {x})^2} = \frac{2\times \dfrac{\sin(x)}{\cos(x)}}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}=\frac{2\times \dfrac{\sin(x)}{\cos(x)}\times \cos^2(x)}{\cos^2(x)-\sin^2(x)}$$
$$=\frac{\sin(x)\cos(x) + \sin(x)\cos(x)}{\cos(x)\cos(x)-\sin(x)\sin(x)}=\frac{\sin(2x)}{\cos(2x)}=\tan(2x)=\sin^2(2x)$$
$$\sin(2x)(1-\sin(2x)\cos(2x))=0$$
So : $ 1-\sin(2x)\cos(2x)=1-\frac 12 \sin(4x)=0 \implies \sin(4x)=2$ which is impossible. 
$\color{red}{\text{or}}$ $\sin(2x)=0\implies x=\dfrac 12 n\pi, n\in \mathbb{Z}$
