= Solution
Take the <imaginary part> of <De Moivre's theorem> at exponent five. Writing $s=\sin\theta$ and $c=\cos\theta$ gives
$$
\sin5\theta=5c^4s-10c^2s^3+s^5
=5s(1-s^2)^2-10s^3(1-s^2)+s^5
=16s^5-20s^3+5s.
$$
For real $x\in[-1,1]$, put $x=\sin\theta$. The equation becomes $\sin5\theta=1$, so $\theta=\pi/10+2\pi k/5$. The distinct sine values are $1$, $(\sqrt5-1)/4$, and $-(1+\sqrt5)/4$. To establish completeness over the complex numbers as well, factor the polynomial:
$$
16x^5-20x^3+5x-1=(x-1)(4x^2+2x-1)^2.
$$
Thus all <polynomial roots>, including their <multiplicities>, are
$$
\boxed{x=1\ \text{(simple)},\qquad
x=\frac{-1+\sqrt5}{4},\ \frac{-1-\sqrt5}{4}\ \text{(each double)}.}
$$
The repeated sine values account for the double roots; there are no additional nonreal roots hidden by the trigonometric parametrization.
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