Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\sin \left( {3x + 1} \right) = \sin \left( {x - 2} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + 1 = x - 2 + k2\pi \\
3x + 1 = \pi  - \left( {x - 2} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - 3 + k2\pi \\
4x = \pi  + 1 + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 3}}{2} + k\pi \\
x = \dfrac{1}{4} + \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
2,\\
\cos \left( {x - \dfrac{\pi }{3}} \right) = \cos \left( {2x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{3} = 2x + \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{3} =  - 2x - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{2} + k2\pi \\
3x = \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
3,\\
\cos 3x = \sin 2x\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - 2x + k2\pi \\
3x = 2x - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
4,\\
\sin \left( {x - 120^\circ } \right) + \cos 2x = 0\\
 \Leftrightarrow  - \sin \left( {x - 120^\circ } \right) = \cos 2x\\
 \Leftrightarrow \sin \left( {120^\circ  - x} \right) = \cos 2x\\
 \Leftrightarrow \cos \left( {90^\circ  - \left( {120^\circ  - x} \right)} \right) = \cos 2x\\
 \Leftrightarrow \cos \left( {x - 30^\circ } \right) = \cos 2x\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 30^\circ  = 2x + k.360^\circ \\
x - 30^\circ  =  - 2x + k.360^\circ 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 30^\circ  + k.360^\circ \\
x = 10^\circ  + k.120^\circ 
\end{array} \right.\\
5,\\
\cos \left( {2x + \dfrac{\pi }{3}} \right) + \cos \left( {x - \dfrac{\pi }{3}} \right) = 0\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) =  - \cos \left( {x - \dfrac{\pi }{3}} \right)\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) = \cos \left[ {\pi  - \left( {x - \dfrac{\pi }{3}} \right)} \right]\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{4\pi }}{3} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = \dfrac{{4\pi }}{3} - x + k2\pi \\
2x + \dfrac{\pi }{3} = x - \dfrac{{4\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{{5\pi }}{3} + k2\pi 
\end{array} \right.
\end{array}\)