Giải phương trình \(\tan (2x + \pi ) = \tan 5x\)
Giải chi tiết:
Điều kiện: \(\left\{ \begin{array}{l}\cos (2x + \pi ) \ne 0\\\cos 5x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x + \pi \ne \frac{\pi }{2} + m\pi \\5x \ne \frac{\pi }{2} + n\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne - \frac{\pi }{4} + \frac{{m\pi }}{2}\\x \ne \frac{\pi }{{10}} + \frac{{n\pi }}{5}\end{array} \right.\;\;\,(m,\;n \in \mathbb{Z})\)
\(\begin{array}{l}\,\,\,\,\,\,\tan (2x + \pi ) = \tan 5x \Leftrightarrow \tan 5x = \tan 2x\\ \Leftrightarrow 5x = 2x + k\pi \Leftrightarrow x = \frac{{k\pi }}{3}\;\;\;\left( {tm} \right)\,\,\,\,\,(k \in Z).\end{array}\)
Chọn B.