Đáp án:

\[\left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x =  \pm \dfrac{\pi }{{12}} + k\pi \\
x =  \pm \dfrac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 ĐKXĐ: \(\left\{ \begin{array}{l}
\cos 2x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{2} + k\pi 
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x \ne \dfrac{\pi }{2} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\)

Ta có:

\(\begin{array}{l}
{\sin ^2}x = \dfrac{{3{{\tan }^2}2x}}{{16{{\cos }^2}x}}\\
 \Leftrightarrow 16{\sin ^2}x.{\cos ^2}x = 3{\tan ^2}2x\\
 \Leftrightarrow 4.\left( {4{{\sin }^2}x.{{\cos }^2}x} \right) = 3{\tan ^2}2x\\
 \Leftrightarrow 4.{\left( {2\sin x.\cos x} \right)^2} = 3{\tan ^2}2x\\
 \Leftrightarrow 4.{\sin ^2}2x = \dfrac{{3{{\sin }^2}2x}}{{{{\cos }^2}2x}}\\
 \Leftrightarrow 4{\sin ^2}2x.{\cos ^2}2x = 3{\sin ^2}2x\\
 \Leftrightarrow 4{\sin ^2}2x.{\cos ^2}2x - 3{\sin ^2}2x = 0\\
 \Leftrightarrow {\sin ^2}2x.\left( {4{{\cos }^2}2x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}2x = 0\\
4{\cos ^2}2x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
{\cos ^2}2x = \dfrac{3}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\cos 2x = \dfrac{{\sqrt 3 }}{2}\\
\cos 2x =  - \dfrac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
2x =  \pm \dfrac{\pi }{6} + k2\pi \\
2x =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x =  \pm \dfrac{\pi }{{12}} + k\pi \\
x =  \pm \dfrac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)