= Solution
Use $z=e^{i\theta}$, so $\sin\theta=(z-z^{-1})/(2i)$ and $d\theta=dz/(iz)$. The <unit circle> is traversed anticlockwise, and the integral becomes
$$
\oint_{|z|=1}\frac{2\,dz}{-3z^2+10iz+3}
=\oint_{|z|=1}\frac{2\,dz}{-3(z-i/3)(z-3i)}.
$$
Only the <simple pole> $z=i/3$ lies inside the <contour>. Its <residue> is
$$
\frac{2}{-6(i/3)+10i}=\frac1{4i}.
$$
Hence the <residue theorem> gives
$$
\boxed{\int_0^{2\pi}\frac{d\theta}{5-3\sin\theta}=2\pi i\frac1{4i}=\frac\pi2}.
$$
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