$\frac{5+i\sin \theta }{5-3i\sin \theta }$  is a real number when

$\frac{5+i\sin \theta }{5-3i\sin \theta }$ is a real number

$\iff \frac{5+i\sin \theta }{5-3i\sin \theta }=\frac{5-i\sin \theta }{5+3i\sin \theta }$

$\iff (5+i\sin \theta )(5+3i\sin \theta )$

$=(5-3i\sin \theta )(5-i\sin \theta )$

$=25-20i\sin \theta -3{\sin }^2\theta$

$\iff 40i\sin \theta =0\iff \sin \theta =0$

This is possible when $\theta =-\mathrm{\pi }$