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