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.

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}
