Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{k\pi }}{4}\\
x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\
x = - \frac{\pi }{{24}} + \frac{{k\pi }}{2}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\[\begin{array}{l}
4\sqrt 3 \cos x.\sin x.\cos 2x = \sin 8x\\
\Leftrightarrow 2\sqrt 3 \sin 2x\cos 2x = 2\sin 4x\cos 4x\\
\Leftrightarrow \sqrt 3 \sin 4x - 2\sin 4x\cos 4x = 0\\
\Leftrightarrow \sin 4x\left( {\sqrt 3 - 2\cos 4x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = 0\\
\cos 4x = \frac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = k\pi \\
4x = \frac{\pi }{6} + k2\pi \\
4x = - \frac{\pi }{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{4}\\
x = \frac{\pi }{{24}} + \frac{{k\pi }}{2}\\
x = - \frac{\pi }{{24}} + \frac{{k\pi }}{2}
\end{array} \right.\,\,\,\left( {k \in Z} \right).
\end{array}\]
