If $\cos x = -\frac{3}{5}$ and $x$ lies in the third quadrant,find the values of the other five trigonometric functions.

Given $\cos x = -\frac{3}{5}$.
Since $\sec x = \frac{1}{\cos x}$,we have $\sec x = -\frac{5}{3}$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Thus,$\sin x = \pm \frac{4}{5}$.
Since $x$ lies in the third quadrant,$\sin x$ must be negative,so $\sin x = -\frac{4}{5}$.
Consequently,$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$.
Now,$\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
Finally,$\cot x = \frac{1}{\tan x} = \frac{3}{4}$.