Solve the following

Solution:

(a) $\pi$

 Given:

$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number

On rationalising, we get,

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

$=\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1)^{2}-(2 i \sin \theta)^{2}}$

$=\frac{3+2 i \sin \theta+6 i \sin \theta+4 i^{2} \sin ^{2} \theta}{1+4 \sin ^{2} \theta}$

$=\frac{3-4 \sin ^{2} \theta+8 i \sin \theta}{1+4 \sin ^{2} \theta} \quad\left[\because i^{2}=-1\right]$

$=\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+i \frac{8 \sin \theta}{1+4 \sin ^{2} \theta}$

For the above term to be real, the imaginary part has to be zero.

$\therefore \frac{8 \sin \theta}{1+4 \sin ^{2} \theta}=0$

$\Rightarrow 8 \sin \theta=0$

For this to be zero,

$\sin \theta=0$

$\Rightarrow \theta=0, \pi, 2 \pi, 3 \pi \ldots$

But $0<\theta<2 \pi$

Hence, $\theta=\pi$