[tex]\mathrm{tg}\Big(x+\dfrac{\pi}{3}\Big) = \mathrm{tg}\Big(\dfrac{\pi}{2}-x\Big) \\ \\ \\x+\dfrac{\pi}{3} = k\pi+\Big(\dfrac{\pi}{2}-x\Big),\quad k\in \mathbb{Z}[/tex]
[tex]\boxed{1}\quad k = 0\Rightarrow x+\dfrac{\pi}{3} =\dfrac{\pi}{2}-x \Rightarrow x =\dfrac{\pi}{12} \\ \boxed{2}\quad k=-1 \Rightarrow x+\dfrac{\pi}{3} = -\pi+\dfrac{\pi}{2}-x\Rightarrow x < 0 \\ \boxed{3}\quad k = 1 \Rightarrow x+\dfrac{\pi}{3}=\pi+\dfrac{\pi}{2}-x \Rightarrow x = \dfrac{7\pi}{12} \\\\ \boxed{4}\quad k=2 \Rightarrow x +\dfrac{\pi}{3}= 2\pi+\dfrac{\pi}{2}-x \Rightarrow x = \dfrac{13\pi}{12} > \pi \\ \\ \\ \Rightarrow \boxed{x\in \Big\{\dfrac{\pi}{12},\,\dfrac{7\pi}{12}\Big\}}[/tex]
