Đáp án:
\(x = \dfrac{{7\pi }}{{12}} + k\pi \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
\sin \left( {2x - \dfrac{\pi }{4}} \right) \ne 0\\
\sin \left( {x + \dfrac{\pi }{3}} \right) \ne 0
\end{array} \right. \to \left\{ \begin{array}{l}
2x - \dfrac{\pi }{4} \ne k\pi \\
x + \dfrac{\pi }{3} \ne k\pi
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x \ne - \dfrac{\pi }{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
\cot \left( {2x - \dfrac{\pi }{4}} \right) = \cot \left( {x + \dfrac{\pi }{3}} \right)\\
\to \tan \left( {2x - \dfrac{\pi }{4}} \right) = \tan \left( {x + \dfrac{\pi }{3}} \right)\\
\to 2x - \dfrac{\pi }{4} = x + \dfrac{\pi }{3} + k\pi \\
\to x = \dfrac{{7\pi }}{{12}} + k\pi \left( {k \in Z} \right)
\end{array}\)
