Đáp án:

4) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\sin \left( {2x + \dfrac{\pi }{6}} \right) = \sin x\\
 \to \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = x + k2\pi \\
2x + \dfrac{\pi }{6} = \pi  - x + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
3)\sin \left( {2x - \dfrac{\pi }{4}} \right) + \sin x = 0\\
 \to 2\sin \left( {\dfrac{{2x - \dfrac{\pi }{4} + x}}{2}} \right).\cos \left( {\dfrac{{2x - \dfrac{\pi }{4} - x}}{2}} \right) = 0\\
 \to \sin \left( {\dfrac{{3x - \dfrac{\pi }{4}}}{2}} \right).\cos \left( {\dfrac{{x - \dfrac{\pi }{4}}}{2}} \right) = 0\\
 \to \left[ \begin{array}{l}
\sin \left( {\dfrac{{3x}}{2} - \dfrac{\pi }{8}} \right) = 0\\
\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{3x}}{2} - \dfrac{\pi }{8} = k\pi \\
\dfrac{x}{2} - \dfrac{\pi }{8} = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{4} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
2)\sin \left( {3x - \dfrac{\pi }{4}} \right) = \sin x\\
 \to \left[ \begin{array}{l}
3x - \dfrac{\pi }{4} = x + k2\pi \\
3x - \dfrac{\pi }{4} = \pi  - x + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{4} + k2\pi \\
4x = \dfrac{{5\pi }}{4} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + k\pi \\
x = \dfrac{{5\pi }}{{16}} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
4)\cos \left( {3x - \dfrac{\pi }{6}} \right) = \cos 2x\\
 \to \left[ \begin{array}{l}
3x - \dfrac{\pi }{6} = 2x + k2\pi \\
3x - \dfrac{\pi }{6} =  - 2x + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
5x = \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)