Ta có:
\(\begin{array}{l}\tan \left( {\frac{\pi }{4} - x} \right) = \tan 2x\\DK:\,\,\left\{ \begin{array}{l}\frac{\pi }{4} - x \ne \frac{\pi }{2} + m\pi \\2x \ne \frac{\pi }{2} + m\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x \ne - \frac{\pi }{4} - m\pi \\x \ne \frac{\pi }{4} + \frac{{m\pi }}{2}\end{array} \right.\\ \Leftrightarrow x \ne \frac{\pi }{4} + \frac{{m\pi }}{2}\,\,\left( {m \in Z} \right)\end{array}\)
Khi đó phương trình tương đương với:
\(\begin{array}{l}\,\,\,\,\,\,\,2x = \frac{\pi }{4} - x + k\pi \\ \Leftrightarrow 3x = \frac{\pi }{4} + k\pi \\ \Leftrightarrow x = \frac{\pi }{{12}} + \frac{{k\pi }}{3}\,\,\,\left( {k \in Z} \right)\end{array}\)
Kết hợp điều kiện ta có:
\(\begin{array}{l}\,\,\,\,\,\,\frac{\pi }{{12}} + \frac{{k\pi }}{3} \ne \frac{\pi }{4} + \frac{{m\pi }}{2}\\ \Leftrightarrow \frac{{k\pi }}{3} \ne \frac{{m\pi }}{2} + \frac{\pi }{6}\\ \Leftrightarrow k \ne \frac{{3m + 1}}{2}\,\,\,\left( {k,m \in Z} \right)\end{array}\)
Vậy phương trình có nghiệm: \(x = \frac{\pi }{{12}} + \frac{{k\pi }}{3}\,\,\,\left( {k \ne \frac{{3m + 1}}{2}\,\,\,\left( {k,m \in Z} \right)} \right)\)
