Question

Find real $\theta$ such that $\frac{3+2i\sin \theta }{1-2i\sin \theta }$, is purely Real.

Answer 1

$\Rightarrow \frac{3+2i\sin \theta }{1-2i\sin \theta }$

$\Rightarrow \frac{\left(3+2i\sin \theta \right)\left(1+2i\sin \theta \right)}{\left(1-2i\sin \theta \right)\left(1+2i\sin \theta \right)}$

$\Rightarrow \frac{\left(3-4{\sin }^2\theta \right)\left(6\sin \theta +2\sin \theta \right)i}{\left(1-4{\sin }^2\theta \right)}$

For purely real,

$\Rightarrow \frac{\left(6\sin \theta +2\sin \theta \right)}{\left(1-4{\sin }^2\theta \right)}=0$

$\Rightarrow 8\sin \theta =0$

$\Rightarrow \sin \theta =0$

$\Rightarrow \theta =0,\pi ,2\pi$