Let $z = \cos(\frac{\pi k}{5}) + i\sin(\frac{\pi k}{5})$
Consider the imaginary part of $z^5$, and deduce that $x^4 - 3x^2 + 1 = 0$ has solutions:
$2\cos(\frac{\pi}{5}), ~2\cos(\frac{2\pi}{5}), ~2\cos(\frac{3\pi}{5}), ~2\cos(\frac{4\pi}{5})$
So, 'considering' the imaginary part of $z$, I considered the following to be true:
$5\cos^4\theta~\sin\theta - 10\cos^2\theta~\sin^3\theta + \sin^5\theta = 0$ where $\theta = \frac{\pi k}{5}$ since $z^5$ is real $\forall_k \in \mathbb{Z}$
How do I apply this to the polynomial from the question? Normally I would have just used the quadratic formula to solve it, but if I do that I get:
$x = \pm\sqrt{\frac{3 \pm \sqrt{5}}{2}}$
but I couldn't seem to relate the two sets of answers easily.