Đáp án:
$\begin{array}{l}
j)\sin x.\left( {\sin 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
k)\sin \left( {x - {{120}^0}} \right) - \cos 2x = 0\\
\Leftrightarrow \sin \left( {x - {{120}^0}} \right) = \cos 2x\\
\Leftrightarrow \sin \left( {x - {{120}^0}} \right) = \sin \left( {{{90}^0} - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - {120^0} = {90^0} - 2x + k{.360^0}\\
x - {120^0} = {180^0} - {90^0} + 2x + k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = {70^0} + k{.120^0}\\
x = - {210^0} - k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = {70^0} + k{.120^0}\\
x = - {210^0} - k{.360^0}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
