Đáp án:

\[\left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x =  - \dfrac{{5\pi }}{{36}} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\sin x =  - \sin \left( { - x} \right)\\
\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\\
\cos x = \cos y \Leftrightarrow \left[ \begin{array}{l}
x = y + k2\pi \\
x =  - y + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {5x + \dfrac{\pi }{6}} \right) = 0\\
 \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) =  - \sin \left( {x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \sin \left( { - x - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \cos \left[ {\dfrac{\pi }{2} - \left( { - x - \dfrac{\pi }{6}} \right)} \right]\\
 \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{{2\pi }}{3} + x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
5x + \dfrac{\pi }{6} = \dfrac{{2\pi }}{3} + x + k2\pi \\
5x + \dfrac{\pi }{6} =  - \dfrac{{2\pi }}{3} - x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x - x = \dfrac{{2\pi }}{3} - \dfrac{\pi }{6} + k2\pi \\
5x + x =  - \dfrac{{2\pi }}{3} - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k2\pi \\
6x =  - \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x =  - \dfrac{{5\pi }}{{36}} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)