Prove that $\csc 2x + \cot 2x = \cot x$. Hence, solve $\csc x + \cot x = \sqrt{3}$ where $0 \le x \le 2\pi$.

1. Express the left-hand side (LHS) in terms of sine and cosine.

$\text{LHS} = \csc 2x + \cot 2x$
$= \frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x}$

2. Combine the terms into a single fraction.

$= \frac{1 + \cos 2x}{\sin 2x}$

3. Apply double angle formulae to express the numerator and denominator in terms of single angles.

Using $\cos 2x = 2\cos^2 x - 1$ and $\sin 2x = 2\sin x \cos x$:
$= \frac{1 + (2\cos^2 x - 1)}{2\sin x \cos x}$

4. Simplify the expression to obtain the right-hand side (RHS).

$= \frac{2\cos^2 x}{2\sin x \cos x}$
$= \frac{\cos x}{\sin x}$
$= \cot x$
$= \text{RHS (shown)}$