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.
 A: since
$$\dfrac{\tan^2{x}\sin^2{x}}{1-\sin^2{x}\cos{2x}}=\dfrac{\tan^2{x}}{\csc^2{x}-\cos{2x}}=\dfrac{\tan^2{x}}{\cot^2{x}+2\sin^2{x}}$$
and simaler other 
$$\dfrac{\cot^2{x}\cos^2{x}}{1-\cos^2{x}\cos{2x}}=\dfrac{\cot^2{x}}{\sec^2{x}-\cos{2x}}=\dfrac{\cot^2{x}}{\tan^2{x}+2\sin^2{x}}$$
so your equation is
$$\dfrac{\tan^2{x}}{\cot^2{x}+2\sin^2{x}}+\dfrac{\cot^2{x}}{\tan^2{x}+2\sin^2{x}}+\dfrac{2\sin^2{x}}{\tan^2{x}+\cot^2{x}}=\dfrac{3}{2}$$
let
$$a=\tan^2{x},b=\cot^2{x},c=2\sin^2{x}$$
then
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{3}{2}$$
But we known well inequality
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\tag{1}$$
$=$ iff$a=b=c$
so $\tan^2{x}=\cot^2{x}=2\sin^2{x}$
ADD $(1)$ proof
By Cauchy-Schwarz inequality  we have
$$\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a(b+c)+b(c+a)+c(a+b)\right)\ge(a+b+c)^2$$
it is enought to prove
$$(a+b+c)^2\ge 3(ab+bc+ac)$$ it is clear
$"="$ iff $a=b=c$
A: 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.
A: For $$\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}$$
it can be seen that the values of $x$ that satisfy this result are $4 x_{n} = (2n+1) \pi$, $n \geq 0$. For each value of $x_{n}$ the values $\sin x_{n} = \pm \frac{1}{\sqrt{2}}$ and $\cos x_{n} = \pm \frac{1}{\sqrt{2}}$. Now, 
\begin{align}
\phi_{n} &= \frac{\tan^2 x_{n} \, \sin^2 x_{n}}{1-\sin^2 x_{n} \,  \cos(2x_{n})}+\frac{\cot^2 x_{n} \, \cos^2 x_{n}}{1-\cos^2 x_{n} \, \cos(2x_{n})}+\frac{2\sin^2 x_{n}}{\tan^2 x_{n} + \cot^2 x_{n}} \\
&= \frac{\frac{1}{2}}{1 - \frac{1}{2} \cdot 0} + \frac{\frac{1}{2}}{1 - \frac{1}{2} \cdot 0} + \frac{2 \cdot \frac{1}{2}}{1 + 1} \\
&= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}.
\end{align} 
Hence 
\begin{align}
x_{n} = \frac{(2n+1) \, \pi}{4} \hspace{5mm} n \geq 0.
\end{align}
