Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{\sin ^2}\left( {3x - \dfrac{{2\pi }}{3}} \right) - {\cos ^2}\left( {\dfrac{x}{3} + \dfrac{\pi }{6}} \right) = 0\\
 \Leftrightarrow {\sin ^2}\left( {3x - \dfrac{{2\pi }}{3}} \right) - {\sin ^2}\left[ {\dfrac{\pi }{2} - \left( {\dfrac{x}{3} + \dfrac{\pi }{6}} \right)} \right] = 0\\
 \Leftrightarrow {\sin ^2}\left( {3x - \dfrac{{2\pi }}{3}} \right) - {\sin ^2}\left( {\dfrac{\pi }{3} - \dfrac{x}{3}} \right) = 0\\
 \Leftrightarrow {\sin ^2}\left( {3x - \dfrac{{2\pi }}{3}} \right) = {\sin ^2}\left( {\dfrac{\pi }{3} - \dfrac{x}{3}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {3x - \dfrac{{2\pi }}{3}} \right) = \sin \left( {\dfrac{\pi }{3} - \dfrac{x}{3}} \right)\\
\sin \left( {3x - \dfrac{{2\pi }}{3}} \right) =  - \sin \left( {\dfrac{\pi }{3} - \dfrac{x}{3}} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {3x - \dfrac{{2\pi }}{3}} \right) = \sin \left( {\dfrac{\pi }{3} - \dfrac{x}{3}} \right)\\
\sin \left( {3x - \dfrac{{2\pi }}{3}} \right) = \sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{3}} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{{2\pi }}{3} = \dfrac{\pi }{3} - \dfrac{x}{3} + k2\pi \\
3x - \dfrac{{2\pi }}{3} = \dfrac{{2\pi }}{3} + \dfrac{x}{3} + k2\pi \\
3x - \dfrac{{2\pi }}{3} = \dfrac{x}{3} - \dfrac{\pi }{3} + k2\pi \\
3x - \dfrac{{2\pi }}{3} = \dfrac{{4\pi }}{3} - \dfrac{x}{3} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{10}} + \dfrac{{3k\pi }}{5}\\
x = \dfrac{\pi }{2} + \dfrac{{3k\pi }}{4}\\
x = \dfrac{\pi }{8} + \dfrac{{3k\pi }}{4}\\
x = \dfrac{{3\pi }}{5} + \dfrac{{3k\pi }}{5}
\end{array} \right.
\end{array}\)