Прошууу у срочно надо даю 50 баллов

Решите задачу:

$\displaystyle tg^2x-(\sqrt{3}-1)tgx-\sqrt{3}=0\\\\tgx=t\\\\t^2-(\sqrt{3}-1)t-\sqrt{3}=0\\\\D=(\sqrt{3}-1)^2-4*(-\sqrt{3})=(3-2\sqrt{3}+1)+4\sqrt{3}=\\\\=3+2\sqrt{3}+1=(\sqrt{3}+1)^2\\\\t_{1.2}=\frac{(\sqrt{3}-1)\pm(\sqrt{3}+1)}{2}\\\\t_1=\frac{\sqrt{3}-1+\sqrt{3}+1}{2}=\sqrt{3}\\\\t_2=\frac{\sqrt{3}-1-\sqrt{3}-1}{2}=-1\\\\tgx=\sqrt{3}; x=\frac{\pi }{3}+\pi n; n\in Z\\\\tgx=-1; x=-\frac{\pi }{4}+\pi n; n\in Z$