= Solution
The complex sine has simple zeros exactly at $z=k\pi$, $k\in\mathbb Z$, because $\sin(k\pi)=0$ and $\cos(k\pi)=(-1)^k\neq0$. Hence the denominator has a zero of order three at each of these points and nowhere else.
At $0$, the numerator has order two, so the quotient has a <simple pole>. Indeed its leading term is $-\pi^4/z$. At $\pi$, the numerator has order one, so there is a <pole> of order two. At $-\pi$, the numerator has order three, cancelling the denominator's order-three zero, so the singularity is removable. Its extension value is
$$
\lim_{z\to-\pi}\frac{z^2(z-\pi)(z+\pi)^3}{\sin^3z}
=\frac{(-\pi)^2(-2\pi)}{\cos^3(-\pi)}=2\pi^3.
$$
At every remaining $k\pi$, the numerator is nonzero, giving a <pole> of order three. Thus the complete finite-plane classification is
$$
\boxed{\begin{array}{c|c}
z&\text{classification}\\\hline
0&\text{pole of order }1\\
\pi&\text{pole of order }2\\
-\pi&\text{removable, extension value }2\pi^3\\
k\pi,\quad k\in\mathbb Z\setminus\{-1,0,1\}&\text{pole of order }3
\end{array}}
$$
Away from this discrete set the denominator is nonzero, so the quotient is holomorphic.
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