Prove the identity $\frac{\sin 2A + \cos A}{1 - \cos 2A + \sin A} = \cot A$ Hence, solve, for $-\pi \le x \le \pi$, the equation $\frac{1 - \cos 4x + \sin 2x}{\sin 4x + \cos 2x} = 5 - 2\sec^2 2x$ giving your answers correct to two decimal places.

1. Apply double angle formulae to the numerator and denominator

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

2. Simplify the denominator

$\text{LHS} = \frac{2\sin A \cos A + \cos A}{2\sin^2 A + \sin A}$

3. Factorise numerator and denominator, then cancel common factors to prove the identity

$\text{LHS} = \frac{\cos A (2\sin A + 1)}{\sin A (2\sin A + 1)}$
$\text{LHS} = \frac{\cos A}{\sin A}$
$\text{LHS} = \cot A = \text{RHS}$

1. Use the identity from part (a) to simplify the left-hand side of the equation.

From part (a), $\frac{\sin 2A + \cos A}{1 - \cos 2A + \sin A} = \cot A$.
Let $A = 2x$. The left-hand side of the given equation is the reciprocal of the left-hand side of the identity in (a).
Therefore, $\frac{1 - \cos 4x + \sin 2x}{\sin 4x + \cos 2x} = \frac{1}{\cot 2x} = \tan 2x$.

2. Substitute into the equation and use the Pythagorean identity to form a quadratic equation in $\tan 2x$.

The equation becomes:
$\tan 2x = 5 - 2\sec^2 2x$
Using the identity $\sec^2 2x = 1 + \tan^2 2x$:
$\tan 2x = 5 - 2(1 + \tan^2 2x)$
$\tan 2x = 5 - 2 - 2\tan^2 2x$
$2\tan^2 2x + \tan 2x - 3 = 0$

3. Solve the quadratic equation for $\tan 2x$.

Factorising the quadratic equation:
$(2\tan 2x + 3)(\tan 2x - 1) = 0$
$\tan 2x = -1.5$ or $\tan 2x = 1$

4. Find the basic angles and solve for $x$ in the given range.

For $-\pi \le x \le \pi$, the range for $2x$ is $-2\pi \le 2x \le 2\pi$.

Case 1: $\tan 2x = 1$
Basic angle $\alpha = \frac{\pi}{4}$
$2x = -\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}$
$x = -\frac{7\pi}{8}, -\frac{3\pi}{8}, \frac{\pi}{8}, \frac{5\pi}{8}$
$x \approx -2.75, -1.18, 0.39, 1.96$

Case 2: $\tan 2x = -1.5$
Basic angle $\alpha = \tan^{-1}(1.5) \approx 0.98279$
$2x = -\pi - 0.98279, -0.98279, \pi - 0.98279, 2\pi - 0.98279$
$2x \approx -4.1244, -0.9828, 2.1588, 5.3004$
$x \approx -2.06, -0.49, 1.08, 2.65$