$\begin{array}{l}1)\,\,\tan(2x -1) = \sqrt3 \quad (1)\\ ĐKXĐ: \, \cos(2x-1) \ne 0 \Leftrightarrow x \ne \dfrac{1}{2} + \dfrac{\pi}{4} + n\dfrac{\pi}{2} \quad (n \in \Bbb Z)\\ (1) \Leftrightarrow \tan(2x -1) = \tan\dfrac{\pi}{3}\\ \Leftrightarrow 2x - 1 = \dfrac{\pi}{3} +k\pi\\ \Leftrightarrow 2x = 1 + \dfrac{\pi}{3} +k\pi\\ \Leftrightarrow x = \dfrac{1}{2} + \dfrac{\pi}{6} +k\dfrac{\pi}{2}\quad (k \in \Bbb Z)\\ 2)\,\,\cot2x = \cot\left(-\dfrac{1}{3}\right)\quad (2)\\ ĐKXĐ: \, \sin2x \ne 0 \Leftrightarrow x \ne n\dfrac{\pi}{2} \quad (n \in \Bbb Z)\\ (2) \Leftrightarrow 2x = - \dfrac{1}{3} +k\pi\\ \Leftrightarrow x = -\dfrac{1}{6} +k\dfrac{\pi}{2}\\ 3)\,\,\cot\left(\dfrac{x}{4} +20^o\right) = -\sqrt3\quad (3)\\ ĐKXĐ: \,\sin\left(\dfrac{x}{4} +20^o\right) \ne 0 \Leftrightarrow x \ne -20^o + n.720^o \quad (n\in\Bbb Z)\\ (3) \Leftrightarrow \cot\left(\dfrac{x}{4} +20^o\right) = \cot(-30^o)\\ \Leftrightarrow \dfrac{x}{4} + 20^o = -30^o + k.180^o\\ \Leftrightarrow \dfrac{x}{4} = -50^o + k.180^o\\ \Leftrightarrow x = -200^o + k.720^o\\ 4)\,\,\cot3x = \tan\dfrac{2\pi}{5}\quad (4)\\ ĐKXĐ: \, \sin3x \ne 0 \Leftrightarrow x \ne n\dfrac{\pi}{3} \quad (n \in \Bbb Z)\\ (4) \Leftrightarrow \cot3x = \cot\left(\dfrac{\pi}{2} - \dfrac{2\pi}{5}\right)\\ \Leftrightarrow 3x = \dfrac{\pi}{10} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{30} + k\dfrac{\pi}{3}\\ 5)\,\,\tan2x = \tan\left(x + \dfrac{\pi}{4}\right) \quad (5)\\ ĐKXĐ:\,\cos2x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{4} +n\dfrac{\pi}{2} \quad (n \in \Bbb Z)\\(5) \Leftrightarrow 2x = x + \dfrac{\pi}{4} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{4} +k\pi \quad (k \in \Bbb Z)\\ 6)\,\,2\tan5x = 1\quad (6)\\ ĐKXĐ: \, \cos5x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{10} + n\dfrac{\pi}{5} \quad (n \in \Bbb Z)\\ (6) \Leftrightarrow \tan5x = \dfrac{1}{2}\\ \Leftrightarrow 5x = \arctan\dfrac{1}{2} +k\pi\\ \Leftrightarrow x = \dfrac{1}{5}\arctan\dfrac{1}{2} +k\dfrac{\pi}{5} \quad (k \in \Bbb Z) \end{array}$