Find the value of $\cos(2\pi /5)$ using radicals This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about $\cos(2\pi/7)$, which I think is a much harder problem.
I dont have the faintest idea how to solve this and the textbook (Hungerford) doesn't have any examples at all. Ive tried looking for resources online but havent found any that I was able to understand.
So can anyone show me how to solve these types of problems? Thanks a bunch.

Note that
$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$
therefore
$\cos(2⋅\dfrac{2\pi}{5})=\cos(3⋅\dfrac{2\pi}{5})$.
Put $\cos(\dfrac{2\pi}{5})=x$. Using the formulas
$\cos2x=2\cos2x−1,\cos 3x=4\cos 3x−3\cos x$,
we have
$4x^3−2x^2−3x+1=0⇔(x−1)(4x^2+2x−1)=0$.
Because $\cos(\dfrac{2\pi}{5})≠1$, we get
$4x^2+2x−1=0.$
Another way, $\cos(\dfrac{2\pi}{5})>0$, then $\cos \dfrac{2\pi}{5}  = \dfrac{-1 + \sqrt{5}}{2}$.
