sin(pi+pi/6)=cos(pi/3-2x) ai giúp em với ạ!

Đáp án:

$\begin{array}{l}
\sin \left( {x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{3} - 2x} \right)\\
 \Rightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{3} + 2x} \right)\\
 \Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{6} = \dfrac{\pi }{2} - \dfrac{\pi }{3} + 2x + k2\pi \\
x + \dfrac{\pi }{6} = \pi  - \left( {\dfrac{\pi }{2} - \dfrac{\pi }{3} + 2x} \right) + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  - k2\pi \\
3x = \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}$