Тригонометрическое уравнение, помогите: tg3x=(2+sqrt3)tgx

Решите задачу:

$tg3x=(2+\sqrt3)tgx\\\\ \frac{3tgx-tg^3x}{1-3tg^2x} =(2+\sqrt3)tgx\\\\t=tgx\; \; \to \; \; \frac{3t-t^3}{1-3t^2} =(2+\sqrt3)t\\\\3t-t^3=(2+\sqrt3)\cdot t\cdot (1-3t^2)\\\\3t-t^3=(2+\sqrt3)\cdot t-3(2+\sqrt3)t^3\\\\(3-2+\sqrt3)\cdot t+(6+3\sqrt3-1)t^3=0\\\\t\cdot ((1+\sqrt3)+(5+3\sqrt3)t^2)=0\\\\t\cdot (1+\sqrt3+(5+3\sqrt3)t^2)=0\\\\Tak\; kak\; \; 1+\sqrt3+(5+3\sqrt3)t^2\ \textgreater \ 0\; ,\; to\; \; mnozitel\; \; t=0\; .\\\\tgx=0\; \; \to \; \; x=\pi n,\; n\in Z$