Đáp án:$\left[ \begin{array}{l}
x = \arctan \left( 5 \right) + k\pi \\
x = \frac{{ - \pi }}{4} + k\pi
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
{\rm{tanx - 5cotx = 4}}\\
{\rm{DKXD:}}\left\{ \begin{array}{l}
{\mathop{\rm s}\nolimits} {\rm{inx}} \ne 0\\
{\rm{cosx}} \ne {\rm{0}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne k\pi \\
x \ne \frac{\pi }{2} + k\pi
\end{array} \right. \Rightarrow x \ne \frac{{k\pi }}{2}\\
pt \Rightarrow {\rm{tanx - }}\frac{5}{{{\rm{tanx}}}} = 4\\
\Rightarrow {\rm{ta}}{{\rm{n}}^2}{\rm{x - 4tanx}} - 5 = 0\\
\Rightarrow \left[ \begin{array}{l}
{\rm{tanx = 5}}\\
{\rm{tanx = - 1}}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \arctan \left( 5 \right) + k\pi \\
x = \frac{{ - \pi }}{4} + k\pi
\end{array} \right.
\end{array}$
