1) $\tan\left(5x - \dfrac{7}{5}\right) - \cot\left(x + \dfrac{\pi}{8}\right) = 0$ $(*)$
$ĐK: \, \begin{cases}\cos\left(5x - \dfrac{7}{5}\right) \ne 0\\\sin\left(x + \dfrac{\pi}{8}\right) \ne 0\end{cases}$
$\Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{10} + \dfrac{7}{25} + k\dfrac{\pi}{5}\\x \ne -\dfrac{\pi}{8} + k\pi\end{cases}$
$(*) \Leftrightarrow \tan\left(5x - \dfrac{7}{5}\right) = \cot\left(x + \dfrac{\pi}{8}\right)$
$\Leftrightarrow \tan\left(5x - \dfrac{7}{5}\right) = \tan\left(\dfrac{\pi}{2} - x - \dfrac{\pi}{8}\right)$
$\Leftrightarrow 5x - \dfrac{7}{5} = \dfrac{3\pi}{8} - x + k\pi$
$\Leftrightarrow 6x = \dfrac{3\pi}{8} +\dfrac{7}{5} + k\pi$
$\Leftrightarrow x = \dfrac{\pi}{16} + \dfrac{7}{30} + k\dfrac{\pi}{6} \quad (k \in \Bbb Z)$
2) $\tan3x = \tan x$ $(**)$
$ĐK: \, \begin{cases}\cos3x \ne 0\\\cos x \ne 0\end{cases} \Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{6} + k\dfrac{\pi}{3}\\x \ne \dfrac{\pi}{2} + k\pi\end{cases}$
$(**)\Leftrightarrow 3x = x + k\pi$
$\Leftrightarrow x = k\dfrac{\pi}{2}$
$\Rightarrow x = k\pi \quad (k \in \Bbb Z)$
