Find the general values of $x$ satisfying the trigonometric equation 
Find the general values of $x$ satisfying 
  $$
  \frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}
$$ 

It seems to me just some equality case of an inequality. But I am unable to find the inequality. Thanks.

I would get everything 
in terms of sin and cos,
which I will write as
s and c cause I'm lazy.
$\begin{array}\\
\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}
&=\frac{(s^2/c^2)s^2 }{1-s^2(c^2-s^2)}+\frac{(c^2/s^2) c^2 }{1-c^2 (c^2-s^2)}+\frac{2s^2}{(s^2/c^2)+(c^2/s^2)}\\
&=\frac{s^4 }{c^2-s^2c^2(c^2-s^2)}+\frac{c^4 }{s^2-c^2s^2 (c^2-s^2)}+\frac{2s^4c^2}{s^4+c^4}\\
&=\frac{s^4 }{c^2-s^2c^4+s^4c^2}+\frac{c^4 }{s^2-c^4s^2+c^2s^4}+\frac{2s^4c^2}{s^4+c^4}\\
&=\frac{s^4 }{c^2-s^2c^4+s^4c^2}+\frac{c^4 }{s^2-c^4s^2+c^2s^4}+\frac{2s^4c^2}{s^4+c^4}\\
\end{array}
$
At this point, I gave up,
and went to Wolfy,
which said
x ~~ -5.40621425424778,
0.631651356268605,
0.876072957166250,
...
and a bunch of other roots.
Interestingly,
I was able to paste the
MathJax expression
unedited
and it was interpreted correctly.
