= Solution
Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the <zeros> are exactly
$$
\boxed{z_n=\frac\pi2+n\pi,\qquad n\in\mathbb Z.}
$$
At each point $\cos'(z_n)=-\sin z_n\ne0$, so the cosine has a simple zero. Cubing it gives
$$
\cos^3z=[-\sin z_n]^3(z-z_n)^3+O((z-z_n)^4).
$$
The <order of a zero of a holomorphic function> is therefore three at every zero. There are no finite singularities because this is an <entire function>.
To classify infinity, examine $\cos^3(1/w)$ at $w=0$. The identity $\cos^3z=(3\cos z+\cos3z)/4$ shows that its <Laurent series> contains the nonzero negative-power coefficients
$$
\frac{(-1)^m(3+3^{2m})}{4(2m)!}\,w^{-2m},\qquad m\geq1.
$$
There are infinitely many terms in the <principal part>, the criterion for an <essential singularity>. Therefore $\boxed{\infty\text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial <entire function> at infinity.
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