Đáp án:
\(\left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{4} + \frac{{m\pi }}{2}
\end{array} \right.\,\,\left( {k,\,\,m \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\,\,\,\,\,\,2{\sin ^3}x + \cos 2x = \sin x\\
\Leftrightarrow 2{\sin ^3}x + 1 - 2{\sin ^2}x - \sin x = 0\\
\Leftrightarrow 2{\sin ^3}x - 2{\sin ^2}x - \sin x + 1 = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {2{{\sin }^2}x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \frac{1}{{\sqrt 2 }}\\
\sin x = - \frac{1}{{\sqrt 2 }}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{4} + m2\pi \\
x = \frac{{3\pi }}{4} + m2\pi \\
x = - \frac{\pi }{4} + l2\pi \\
x = \frac{{5\pi }}{4} + l2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{4} + \frac{{m\pi }}{2}
\end{array} \right.\,\,\left( {k,\,\,m \in Z} \right).
\end{array}\)
