giải giúp mình ý đầu thôi, với lại sau dấu bằng là 12+ 1/2 sin y

$\begin{array}{l}
d)ĐK:\sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi  \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\left( {k \in \mathbb Z} \right)\\
VT = {\left( {{{\sin }^2}x + \dfrac{1}{{{{\sin }^2}x}}} \right)^2} + {\left( {{{\cos }^2}x + \dfrac{1}{{{{\cos }^2}x}}} \right)^2}\\
 = {\sin ^4}x + \dfrac{1}{{{{\sin }^4}x}} + 2 + {\cos ^4}x + \dfrac{1}{{{{\cos }^4}x}} + 2\\
 = {\sin ^4}x + \dfrac{1}{{{{\sin }^4}x}} + {\cos ^4}x + \dfrac{1}{{{{\cos }^4}x}} + 4\\
 = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {1 + \dfrac{1}{{{{\sin }^4}x{{\cos }^4}x}}} \right) + 4\\
 = \left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)\left( {1 + \dfrac{{16}}{{{{\sin }^2}2x}}} \right) + 4\\
 \ge \left( {1 - \dfrac{1}{2}.1} \right)\left( {1 + 16} \right) + 4 = \dfrac{{25}}{2}\\
VP = 12 + \dfrac{1}{2}\sin y \le 12 + \dfrac{1}{2} = \dfrac{{25}}{2}
\end{array}$

Vậy dấu bằng xảy ra. Dấu bằng xảy ra khi và chỉ khi:

$\left\{ \begin{array}{l}
{\sin ^2}2x = 1\\
\sin y = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
y = \dfrac{\pi }{2} + m2\pi 
\end{array} \right.\left( {k,m \in \mathbb Z} \right)$