 Solution

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