= Solution
Write $z=e^{i\theta}$. The trigonometric sum formulas give
$$
p(e^{i\theta})=2e^{3i\theta}\cos2\theta,\qquad
\operatorname{Im}p=2\sin3\theta\cos2\theta.
$$
Thus the imaginary part is zero either when $\theta=k\pi/3$, $k=0,\ldots,5$, or when $\theta=\pi/4+k\pi/2$, $k=0,\ldots,3$. These two lists are disjoint. At the latter four points $p=0$. At the former six points the real values, in increasing angular order, are
$$
\boxed{2,\ 1,\ -1,\ -2,\ -1,\ 1}.
$$
Equivalently, $z=1$ maps to $2$, $z=-1$ to $-2$, $z=e^{\pm i\pi/3}$ to $1$, $z=e^{\pm2i\pi/3}$ to $-1$, and the four roots of $z^4=-1$ map to zero. These are all the unit-circle points with real <image>.
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