Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = - \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\cos 4x + \sin 3x\cos x - \cos 3x\sin x = 0\\
\to \cos 4x + \sin \left( {3x - x} \right) = 0\\
\to \cos 4x + \sin 2x = 0\\
\to 1 - 2{\sin ^2}2x + \sin 2x = 0\\
\to \left[ \begin{array}{l}
\sin x = 1\\
\sin x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = - \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
