Answer
$$\left( { - 2\sqrt 3 ,\frac{{5\pi }}{6}} \right){\text{ and }}\left( {2\sqrt 3 ,\frac{{11\pi }}{6}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{We have the rectangular coordinates }}\left( {x,y} \right) = \left( {3, - \sqrt 3 } \right) \cr
& {\text{The polar coordinates are:}} \cr
& r = \sqrt {{x^2} + {y^2}} \cr
& r = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - \sqrt 3 } \right)}^2}} \cr
& r = 2\sqrt 3 \cr
& \tan \theta = \frac{{ - \sqrt 3 }}{3} \cr
& \theta = {\tan ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{3}} \right) + \pi = \frac{{5\pi }}{6} \cr
& \theta = {\tan ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{3}} \right) + 2\pi = \frac{{11\pi }}{6} \cr
& {\text{Therefore, the polar coordinates are:}} \cr
& \left( { - 2\sqrt 3 ,\frac{{5\pi }}{6}} \right){\text{ and }}\left( {2\sqrt 3 ,\frac{{11\pi }}{6}} \right) \cr} $$
