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