Answer
$\cos{x}=-\dfrac{\sqrt{3}}{2}$
$\tan{x}=\dfrac{\sqrt{3}}{3}$
Work Step by Step
Since $x$ lies in the third quadrant, $\cos{x}$ is negative.
Using the identity:
$$\cos{x}=-\sqrt{1-\sin^2{x}}$$
$$\cos{x}=-\sqrt{1-\left(-\dfrac{1}{2} \right)^2}$$
$$\cos{x}=-\dfrac{\sqrt{3}}{2}$$
$\because \tan{x}=\dfrac{\sin{x}}{\cos{x}}$
$\tan{x}=\dfrac{-\dfrac{1}{2}}{-\dfrac{\sqrt{3}}{2}}$
$\tan{x}=\dfrac{\sqrt{3}}{3}$
