Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\cos \left( {2x + \dfrac{\pi }{3}} \right) + \cos \left( {x - \dfrac{\pi }{6}} \right) = 0\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) =  - \cos \left( {x - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) = \cos \left[ {\pi  - \left( {x - \dfrac{\pi }{6}} \right)} \right]\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{7\pi }}{6} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} - x + k2\pi \\
2x + \dfrac{\pi }{3} = x - \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{{3\pi }}{2} + k2\pi 
\end{array} \right.\\
b,\\
\cos \left( {2x - \dfrac{\pi }{4}} \right) + \sin \left( {\dfrac{\pi }{3} - x} \right) = 0\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) =  - \sin \left( {\dfrac{\pi }{3} - x} \right)\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \sin \left( {x - \dfrac{\pi }{3}} \right)\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \cos \left[ {\dfrac{\pi }{2} - \left( {x - \dfrac{\pi }{3}} \right)} \right]\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{5\pi }}{6} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} - x + k2\pi \\
2x - \dfrac{\pi }{4} = x - \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{13\pi }}{{12}} + \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{{7\pi }}{{12}} + k2\pi 
\end{array} \right.\\
c,\\
\tan \left( {2x + \dfrac{\pi }{4}} \right) = \tan x\\
 \Leftrightarrow 2x + \dfrac{\pi }{4} = x + k\pi \\
 \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi 
\end{array}\)