LOTS OF POUNT PLS HELP

Answer:
Option 2
Step-by-step explanation:
Using the quadrstic formula,
[tex]\tan \theta=\frac{-2 \pm \sqrt{2^2 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} \\ \\ =\frac{-2 \pm 4}{2\sqrt{3}} \\ \\ =\frac{-1 \pm 2}{\sqrt{3}} \\ \\ = -\sqrt{3}, \frac{1}{\sqrt{3}}[/tex]
Case 1
[tex]\tan \theta=-\sqrt{3} \implies \theta=\frac{2\pi}{3}, \frac{5\pi}{3}[/tex]
Case 2
[tex]\tan \theta=\frac{1}{\sqrt{3}} \implies \theta=\frac{\pi}{6}, \frac{7\pi}{6}[/tex]