Prove the identity $\cot \theta - 1 = \dfrac{\cos 2\theta}{\sin \theta (\cos \theta + \sin \theta)}$. Hence solve the equation $\dfrac{\cos 4\theta}{\sin 2\theta(\cos 2\theta + \sin 2\theta)} = -\dfrac{1}{2}$ for $-180^\circ \le \theta \le 180^\circ$.

1. Apply the double angle formula for cosine to the numerator of the right-hand side.

$\text{RHS} = \frac{\cos 2\theta}{\sin \theta (\cos \theta + \sin \theta)}$
$= \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta (\cos \theta + \sin \theta)}$

2. Factorise the numerator using the difference of two squares and cancel the common factor.

$= \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\sin \theta (\cos \theta + \sin \theta)}$
$= \frac{\cos \theta - \sin \theta}{\sin \theta}$

3. Split the fraction and use the definition of cotangent to reach the left-hand side.

$= \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\sin \theta}$
$= \cot \theta - 1$
$= \text{LHS (proven)}$

1. Apply the identity from part (a) to simplify the equation

From part (a), we have $\cot \alpha - 1 = \dfrac{\cos 2\alpha}{\sin \alpha (\cos \alpha + \sin \alpha)}$.
Letting $\alpha = 2\theta$, the equation becomes:
$\cot 2\theta - 1 = -\dfrac{1}{2}$

2. Rearrange to find $\tan 2\theta$

$\cot 2\theta = \dfrac{1}{2}$
$\tan 2\theta = 2$

3. Determine the basic angle and the range for $2\theta$

Basic angle $\alpha = \tan^{-1}(2) \approx 63.435^\circ$
Given $-180^\circ \le \theta \le 180^\circ$, the range for $2\theta$ is $-360^\circ \le 2\theta \le 360^\circ$.

4. Find all possible values for $2\theta$ in the required range

Since $\tan 2\theta > 0$, $2\theta$ lies in the 1st and 3rd quadrants.
$2\theta = 63.435^\circ, 63.435^\circ + 180^\circ, 63.435^\circ - 180^\circ, 63.435^\circ - 360^\circ$
$2\theta = 63.435^\circ, 243.435^\circ, -116.565^\circ, -296.565^\circ$

5. Solve for $\theta$ and round to 1 decimal place

$\theta = 31.717^\circ, 121.717^\circ, -58.282^\circ, -148.282^\circ$
$\theta = -148.3^\circ, -58.3^\circ, 31.7^\circ, 121.7^\circ$ (to 1 d.p.)