Ta có: $tanx= tan(x+\pi)$
* $tanx=\sqrt3$
$\Rightarrow x=arctan\sqrt3 + k\pi=\dfrac{\pi}{3}+k\pi$
* $tanx=-\sqrt3$
$\Rightarrow x=arctan(-\sqrt3)+ k\pi =\dfrac{-\pi}{3}+k\pi= \pi-\dfrac{\pi}{3}+k\pi= \dfrac{2\pi}{3}+k\pi$
* Tương tự với $tanx=\pm 1$, $x=\dfrac{-\pi}{4}+k\pi = \dfrac{3\pi}{4}+k\pi$