Prove that $\csc 2x + \cot 2x = \cot x$. Hence, deduce the value of $\cot 15^\circ$ in surd form. Using part (i), solve $\csc 2x + \cot 2x = 6 - 5\tan x$ for $0^\circ \le x \le 360^\circ$.

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

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

2. Combine into a single fraction and apply double angle formulae.

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

3. Simplify the expression to obtain the right-hand side.

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

1. Substitute $x = 15^\circ$ into the identity from part (i)

From part (i), $\cot x = \csc 2x + \cot 2x$.
Substitute $x = 15^\circ$:
$\cot 15^\circ = \csc(2 \times 15^\circ) + \cot(2 \times 15^\circ)$
$\cot 15^\circ = \csc 30^\circ + \cot 30^\circ$

2. Evaluate the trigonometric ratios of $30^\circ$ using exact values

$\csc 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{\frac{1}{2}} = 2$
$\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$
$\therefore \cot 15^\circ = 2 + \sqrt{3}$

1. Use the result from part (i) to simplify the equation.

From part (i), $\csc 2x + \cot 2x = \cot x$.
Substitute this into the given equation:
$\cot x = 6 - 5\tan x$

2. Express the equation entirely in terms of $\tan x$ and form a quadratic equation.

Since $\cot x = \frac{1}{\tan x}$:
$\frac{1}{\tan x} = 6 - 5\tan x$
Multiply throughout by $\tan x$:
$1 = 6\tan x - 5\tan^2 x$
$5\tan^2 x - 6\tan x + 1 = 0$

3. Solve the quadratic equation for $\tan x$.

Factorising the quadratic equation:
$(5\tan x - 1)(\tan x - 1) = 0$
$\tan x = \frac{1}{5} \quad \text{or} \quad \tan x = 1$

4. Find all possible values of $x$ in the given range $0^\circ \le x \le 360^\circ$.

For $\tan x = 1$:
Basic angle $\alpha = 45^\circ$
$x = 45^\circ, 180^\circ + 45^\circ = 225^\circ$

For $\tan x = \frac{1}{5}$:
Basic angle $\alpha = \tan^{-1}(0.2) \approx 11.3099^\circ$
$x = 11.3^\circ, 180^\circ + 11.3099^\circ = 191.3^\circ$ (to 1 d.p.)

Therefore, $x = 11.3^\circ, 45^\circ, 191.3^\circ, 225^\circ$.