Cardano-Vieta applied to $\cos(5\alpha) = 16\cos^5(\alpha) - 20\cos^3(\alpha) + 5 \cos(\alpha)$ I'm given this equality:
$\cos(5\alpha) = 16\cos^5(\alpha) - 20\cos^3(\alpha) + 5 \cos(\alpha)$
Form there, we make the substitution $x = \cos(\alpha)$ and $p = \cos(5\alpha)$ and we obtain the equation:
$16x^5 - 20x^3 + 5x - p = 0$ and it says the 5 roots of these polynomial are:
$\cos(\alpha),\ \cos(\alpha+72),\ \cos(\alpha+144),\ \cos(\alpha+216),\ \cos(\alpha + 288)$. Why does the polynomial have these roots?
Thanks in advance.
 A: Note that $$\cos(5\alpha) = Re\left[e^{5i\alpha}\right] = Re\left[\left(e^{i \alpha}\right)^5\right]$$
and use De Moivre's formula to find the fifth roots of $(e^{i \alpha})^5$ and then take the real part to get the roots you have listed.
A: Turn the problem around the other way.
We are looking for solutions to $16x^5-20x^3+5x-p=0$, where $p$ is some given constant.
Introduce the change of variables $x=\cos\alpha$.
Now the equation reads
$$16\cos^5\alpha-20\cos^3\alpha+5\cos\alpha=p$$
Using the identity $\cos 5\alpha = 16\cos^5\alpha - 20\cos^3\alpha+5\cos\alpha$, this becomes
$$\cos 5\alpha=p$$
Provided that $p$ is between $1$ and $-1$, this can be solved as follows:
$$5\alpha = \cos^{-1}p$$
$$\alpha = \frac{1}{5}\cos^{-1}p$$
So compute $\alpha$ from that formula, and back-substitute it into $x=\cos\alpha$ to get a solution to the original equation.
However, keep in mind that $5\alpha = \cos^{-1}p$ is just one possible solution to the equation $\cos 5\alpha=p$.  You can get more solutions by adding multiples of $360°$.  That is, we could have any of the following:
$$5\alpha = \cos^{-1}p$$
$$5\alpha=\cos^{-1}p+360$$
$$5\alpha=\cos^{-1}p+720$$
$$5\alpha=\cos^{-1}p+1080$$
$$5\alpha=\cos^{-1}p+1440$$
Dividing these each by 5 gives you the five solutions you have in your question.
