Đáp án:
\(x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{{\sin }^2}2x - 2}}{{4{{\cos }^2}x.({\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}x - 1)}} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\
\frac{{{{\sin }^2}2x - 2}}{{ - 4{{\cos }^2}x}} = {\sin ^2}x\\
{\sin ^2}2x - 2 = - 4{\sin ^2}x.{\cos ^2}x = - {\sin ^2}2x\\
2{\sin ^2}2x - 2 = 0\\
{\sin ^2}2x = 1\\
\left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x = - 1
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} + k2\pi \\
2x = \frac{{ - \pi }}{2} + k2\pi
\end{array} \right. \leftrightarrow 2x = \frac{\pi }{2} + k\pi \leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)(tm)
\end{array}\)
