Đáp án: $x = \dfrac{{k\pi }}{2};x = \pm \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\sin 4x + \sin 2x = 0\\
\Leftrightarrow \sin \left[ {2.\left( {2x} \right)} \right] + \sin 2x = 0\\
\Leftrightarrow 2\sin 2x.\cos 2x + \sin 2x = 0\\
\Leftrightarrow \sin 2x.\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
2\cos 2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
\cos 2x = - \dfrac{1}{2}
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
2x = \dfrac{{2\pi }}{3} + k2\pi \\
2x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{3} + k\pi \\
x = - \dfrac{\pi }{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{k\pi }}{2};x = \pm \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)
\end{array}$
