[tex]\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)[/tex]
[tex]\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}[/tex]
[tex]\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}[/tex]
[tex]x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}[/tex]
