Đáp án:
$\left[ \begin{array}{l}
x = \frac{1}{3} + \frac{{k\pi }}{3}\\
x = \frac{{k\pi }}{2}
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\sin \left( {3x - 1} \right).\tan 2x = 0\\
dkxd:\cos 2x \ne 0 \Rightarrow 2x \ne \frac{\pi }{2} + k\pi \Rightarrow x \ne \frac{\pi }{4} + \frac{{k\pi }}{2}\\
pt \Rightarrow \sin \left( {3x - 1} \right).\sin 2x = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin \left( {3x - 1} \right) = 0\\
\sin 2x = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
3x - 1 = k\pi \\
2x = k\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{1}{3} + \frac{{k\pi }}{3}\\
x = \frac{{k\pi }}{2}
\end{array} \right.
\end{array}$
