Find the values of the other five trigonometric functions if $\cot x = \frac{3}{4}$ and $x$ lies in the third quadrant.

Given $\cot x = \frac{3}{4}$.
$\tan x = \frac{1}{\cot x} = \frac{1}{3/4} = \frac{4}{3}$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$1 + (4/3)^2 = \sec^2 x$
$1 + \frac{16}{9} = \sec^2 x$
$\frac{25}{9} = \sec^2 x$
$\sec x = \pm \frac{5}{3}$.
Since $x$ lies in the third quadrant,$\sec x$ is negative:
$\sec x = -\frac{5}{3}$.
$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$.
Since $\tan x = \frac{\sin x}{\cos x}$,we have $\sin x = \tan x \cdot \cos x$:
$\sin x = \left(\frac{4}{3}\right) \cdot \left(-\frac{3}{5}\right) = -\frac{4}{5}$.
$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$.