Giải thích các bước giải:
a,
\[\begin{array}{l}
2\sin 2x\cos 2x + \sqrt 3 \cos 4x + \sqrt 2 = 0\\
\Leftrightarrow \sin 4x + \sqrt 3 \cos 4x = - \sqrt 2 \\
\Leftrightarrow \frac{1}{2}\sin 4x + \frac{{\sqrt 3 }}{2}\cos 4x = \frac{{ - \sqrt 2 }}{2}\\
\Leftrightarrow \sin \left( {4x + \frac{\pi }{3}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
4x + \frac{\pi }{3} = \frac{{ - \pi }}{4} + k2\pi \\
4x + \frac{\pi }{3} = \frac{{5\pi }}{4} + k2\pi
\end{array} \right.
\end{array}\]
b,
\[\begin{array}{l}
4{\sin ^2}x + 3\sqrt 3 \sin 2x - 2{\cos ^2}x = 4\\
\Leftrightarrow 4{\sin ^2}x + 6\sqrt 3 \sin x\cos x - 2{\cos ^2}x = 4{\sin ^2}x + 4{\cos ^2}x\\
\Leftrightarrow \sqrt 3 \sin x\cos x = {\cos ^2}x\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sqrt 3 \sin x - \cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\frac{{\sqrt 3 }}{2}\sin x - \frac{1}{2}\cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin \left( {x - \frac{\pi }{6}} \right) = 0
\end{array} \right.
\end{array}\]
