Find the values of the other five trigonometric functions if $\cos x = -\frac{1}{2}$ and $x$ lies in the third quadrant.

Given $\cos x = -\frac{1}{2}$.
$\sec x = \frac{1}{\cos x} = \frac{1}{(-1/2)} = -2$.
Using the identity $\sin^2 x + \cos^2 x = 1$:
$\sin^2 x = 1 - \cos^2 x = 1 - (-1/2)^2 = 1 - 1/4 = 3/4$.
Since $x$ lies in the third quadrant,$\sin x$ must be negative.
$\sin x = -\frac{\sqrt{3}}{2}$.
$\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$.