Prove the identity $\dfrac{\sin\theta}{1+\cos\theta} = \tan \dfrac{\theta}{2}$. Hence solve the equation $\dfrac{1+\cos\theta}{\sin\theta} = -3$ for $0^\circ \le \theta \le 360^\circ$.

1. Express the numerator using the double angle formula for sine

$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$

2. Express the denominator using the double angle formula for cosine

$\cos\theta = 2\cos^2\frac{\theta}{2} - 1$
$1 + \cos\theta = 1 + \left(2\cos^2\frac{\theta}{2} - 1\right) = 2\cos^2\frac{\theta}{2}$

3. Substitute into the LHS and simplify to obtain the RHS

$\text{LHS} = \dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}$
$= \dfrac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$
$= \tan\frac{\theta}{2} = \text{RHS}$

1. Rewrite the equation using the identity from part (a)

From part (a), $\dfrac{\sin\theta}{1+\cos\theta} = \tan \dfrac{\theta}{2}$.

The given equation is $\dfrac{1+\cos\theta}{\sin\theta} = -3$.

Taking the reciprocal of both sides:
$\dfrac{\sin\theta}{1+\cos\theta} = -\dfrac{1}{3}$

$\tan \dfrac{\theta}{2} = -\dfrac{1}{3}$

2. Find the basic angle and determine the correct quadrant for $\frac{\theta}{2}$

Basic angle $\alpha = \tan^{-1}\left(\dfrac{1}{3}\right) \approx 18.435^\circ$

Since $0^\circ \le \theta \le 360^\circ$, we have $0^\circ \le \dfrac{\theta}{2} \le 180^\circ$.

Since $\tan \dfrac{\theta}{2}$ is negative, $\dfrac{\theta}{2}$ must lie in the second quadrant.

3. Solve for $\theta$

$\dfrac{\theta}{2} = 180^\circ - 18.435^\circ = 161.565^\circ$

$\theta = 323.13^\circ \approx 323.1^\circ$ (to 1 d.p.)