Prove the trigonometric identity: $\dfrac{\csc^{2}A + 2\cot A}{(\cos A + \sin A)^{2}} = \csc^{2}A$ Hence solve the equation: $\csc^{2}A + 2\cot A = 4(\cos A + \sin A)^{2} \quad\text{for }\; 0 \le x \le 360^{\circ}.$ Given that $\dfrac{1 + \sin x + 2\cos x}{1 + 2\sin x + \cos x} = 1$ and $x$ is acute, find the exact value of $\cos x$.

1. Expand the denominator of the LHS

$\text{LHS} = \dfrac{\csc^{2}A + 2\cot A}{\cos^{2}A + 2\sin A\cos A + \sin^{2}A}$

2. Apply Pythagorean identity to the denominator

$\text{LHS} = \dfrac{\csc^{2}A + 2\cot A}{1 + 2\sin A\cos A}$

3. Express the numerator in terms of sine and cosine

$\text{Numerator} = \dfrac{1}{\sin^{2}A} + \dfrac{2\cos A}{\sin A} = \dfrac{1 + 2\sin A\cos A}{\sin^{2}A}$

4. Substitute back into LHS and simplify to reach RHS

$\text{LHS} = \dfrac{\frac{1 + 2\sin A\cos A}{\sin^{2}A}}{1 + 2\sin A\cos A} = \dfrac{1}{\sin^{2}A} = \csc^{2}A = \text{RHS}$

1. Rearrange the equation to use the identity from part (a)(i).

From (a)(i), we know that $\csc^{2}A + 2\cot A = \csc^{2}A(\cos A + \sin A)^{2}$.
Substitute this into the given equation:
$\csc^{2}A(\cos A + \sin A)^{2} = 4(\cos A + \sin A)^{2}$

2. Factorise the equation to find all possible cases.

$\csc^{2}A(\cos A + \sin A)^{2} - 4(\cos A + \sin A)^{2} = 0$
$(\csc^{2}A - 4)(\cos A + \sin A)^{2} = 0$

3. Solve the first factor for $A$.

Case 1: $\csc^{2}A - 4 = 0$
$\csc^{2}A = 4$
$\sin^{2}A = \frac{1}{4}$
$\sin A = \pm \frac{1}{2}$
For $\sin A = \frac{1}{2}$, basic angle $\alpha = 30^{\circ}$.
$A = 30^{\circ}, 180^{\circ} - 30^{\circ} = 150^{\circ}$
For $\sin A = -\frac{1}{2}$, basic angle $\alpha = 30^{\circ}$.
$A = 180^{\circ} + 30^{\circ} = 210^{\circ}, 360^{\circ} - 30^{\circ} = 330^{\circ}$

4. Solve the second factor for $A$.

Case 2: $\cos A + \sin A = 0$
$\sin A = -\cos A$
$\tan A = -1$
Basic angle $\alpha = 45^{\circ}$.
Since $\tan A$ is negative, $A$ is in the 2nd and 4th quadrants.
$A = 180^{\circ} - 45^{\circ} = 135^{\circ}$
$A = 360^{\circ} - 45^{\circ} = 315^{\circ}$

1. Eliminate the denominator and simplify the equation.

$1 + \sin x + 2\cos x = 1 + 2\sin x + \cos x$
$\cos x = \sin x$

2. Use a trigonometric identity to find the angle $x$.

$\frac{\sin x}{\cos x} = 1$
$\tan x = 1$

3. Determine the exact value of $\cos x$ for the acute angle.

Since $x$ is acute, $x = 45^{\circ}$.
$\cos x = \cos 45^{\circ} = \frac{\sqrt{2}}{2}$