Question: Realise the denominator $\frac{3+2i\sin\theta}{1-2i\sin\theta}$ and hence find $\theta$ if the expression is purely imaginary.
I've realised the denominator $\frac{3+8i\sin\theta-4\sin^2\theta}{1+4\sin^2\theta}$ but don't know how to utilise the knowledge that the expression is purely imaginary.
The second expression that you wrote can be written as $$ \frac{3 + 8i\sin(\theta) - 4\sin(\theta)^2}{1+4\sin(\theta)^2} = \underbrace{\frac{3 - 4\sin(\theta)^2}{1+4\sin(\theta)^2}}_{\text{real part}} + i\underbrace{\frac{8\sin(\theta)}{1+4\sin(\theta)^2}}_{\text{imaginary part}}. $$ If the expression is to be purely imaginary, then the real part must be zero. This occurs only when the numerator of the real part is zero, i.e. when $$ 3 - 4\sin(\theta)^2 = 0. $$ Via some manipulation, this becomes $$ \sin(\theta) = \pm \frac{\sqrt{3}}{2} \implies \theta \in \left\{ \pm \frac{\pi}{3} + 2k\pi \ \middle|\ k \in \mathbb{Z} \right\} \cup \left\{ \pm \frac{2\pi}{3} + 2k\pi \ \middle|\ k\in\mathbb{Z} \right\}.$$ where $k$ is an arbitrary integer.