Find the values of the other five trigonometric functions if $\sin x = \frac{3}{5}$ and $x$ lies in the second quadrant.

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