Đáp án:
$\begin{array}{l}
\cot \dfrac{{2x + 1}}{6} = \tan \dfrac{1}{3}\\
\Rightarrow \cot \dfrac{{2x + 1}}{6} = \cot \left( {{{90}^0} - \arctan \dfrac{1}{3}} \right)\\
\Rightarrow \dfrac{{2x + 1}}{6} = {90^0} - \arctan \dfrac{1}{3} + k{.90^0}\\
\Rightarrow 2x = {540^0} - 1 - 6.arctan\dfrac{1}{3} + k{.540^0}\\
\Rightarrow x = {270^0} - \dfrac{1}{2} - 3.\arctan \dfrac{1}{3} + k{.270^0}
\end{array}$
