If a complex number $\mathrm{z}=\frac{4+3 i \sin \theta}{1-2 i \sin \theta}($ where $i=\sqrt{-1})$ is purely real, then the value of $\theta$ is

The complex number $z = \frac{4+3i\sin\theta}{1-2i\sin\theta}$ is purely real if and only if its imaginary part is zero.

Rationalizing the denominator by multiplying numerator and denominator by the conjugate $1+2i\sin\theta$:

$z = \frac{(4+3i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)} = \frac{4 + 8i\sin\theta + 3i\sin\theta + 6i^2\sin^2\theta}{1 - 4i^2\sin^2\theta}$

Substituting $i^2 = -1$:

$z = \frac{4 + 11i\sin\theta - 6\sin^2\theta}{1 + 4\sin^2\theta} = \frac{4 - 6\sin^2\theta}{1 + 4\sin^2\theta} + i\frac{11\sin\theta}{1 + 4\sin^2\theta}$

The imaginary part must vanish for $z$ to be real. Since $1 + 4\sin^2\theta > 0$ for all $\theta$, we require $11\sin\theta = 0$, hence $\sin\theta = 0$.

The general solution is $\theta = n\pi$ for $n \in \mathbb{Z}$, which corresponds to option D.