<s>[INST] 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
 [/INST] $\frac{2\tan x}{1-\tan^2x}=(\sin 2x)^2$
$\tan 2x=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}-\sin^2 2x=0$
$\frac{\sin 2x}{\cos 2x}(1-\sin 2x \cos 2x)=0$
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.
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