Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
\sin \left( {x - 120^\circ } \right) + \cos 2x = 0\\
\Leftrightarrow \sin \left( {x - 120^\circ } \right) = - \cos 2x\\
\Leftrightarrow \sin \left( {x - 120^\circ } \right) = \cos \left( {180^\circ - 2x} \right)\\
\Leftrightarrow \sin \left( {x - 120^\circ } \right) = \sin \left[ {90^\circ - \left( {180^\circ - 2x} \right)} \right]\\
\Leftrightarrow \sin \left( {x - 120^\circ } \right) = \sin \left( {2x - 90^\circ } \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - 120^\circ = 2x - 90^\circ + k.360^\circ \\
x - 120^\circ = 180^\circ - \left( {2x - 90^\circ } \right) + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 30^\circ + k.360^\circ \\
x = 130^\circ + k120^\circ
\end{array} \right.\\
4,\\
\sin 3x + \sin \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = 0\\
\Leftrightarrow \sin 3x = - \sin \left( {\dfrac{\pi }{4} - \dfrac{\pi }{2}} \right)\\
\Leftrightarrow \sin 3x = \sin \left( {\dfrac{x}{2} - \dfrac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{x}{2} - \dfrac{\pi }{4} + k2\pi \\
3x = \pi - \left( {\dfrac{x}{2} - \dfrac{\pi }{4}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{{10}} + \dfrac{{4k\pi }}{5}\\
x = \dfrac{{5\pi }}{{14}} + \dfrac{{4\pi }}{7}
\end{array} \right.
\end{array}\)
