\[\begin{array}{l}
4\left( {{{\sin }^2}x + \frac{1}{{{{\sin }^2}x}}} \right) - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) = 7\,\,\,\,\left( {DK:\,\,\,\sin x \ne 0} \right)\\
\Leftrightarrow 4\left( {{{\sin }^2}x - 2 + \frac{1}{{{{\sin }^2}x}} + 2} \right) - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) = 7\\
\Leftrightarrow 4{\left( {\sin x - \frac{1}{{\sin x}}} \right)^2} + 8 - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) - 7 = 0\\
\Leftrightarrow 4{\left( {\sin x - \frac{1}{{\sin x}}} \right)^2} - 4\left( {\sin x - \frac{1}{{\sin x}}} \right) + 1 = 0\\
\Leftrightarrow {\left[ {2\left( {\sin x - \frac{1}{{\sin x}}} \right) - 1} \right]^2} = 0\\
\Leftrightarrow 2\left( {\sin x - \frac{1}{{\sin x}}} \right) - 1 = 0\\
\Leftrightarrow \sin x - \frac{1}{{\sin x}} = \frac{1}{2}\\
\Leftrightarrow 2{\sin ^2}x - 2 = \sin x\\
\Leftrightarrow 2{\sin ^2}x - \sin x - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \frac{{1 + \sqrt {17} }}{4}\,\,\,\,\left( {ktm} \right)\\
\sin x = \frac{{1 - \sqrt {17} }}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi \,\\
x = \pi - \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).
\end{array}\]
