Trisecting $2\pi/5$, is this possible? I guess that the answer is no, even knowing that $cos(2\pi/5)$ is constructible since the $5$th root o unity is construtctible.
But when I use the trick for finding the minimal polynomial of $3\theta=2\pi/5$ I get that $\theta$ is the root of 
$p(x)=4x^3 - 3x - cos(2\pi/5)$
and this polynomial is not even on $\mathbb{Q}[x]$, so how shoul i proceed to prove that is it or isn't possible to trisect $\theta?$

The minimal polynomial for an $n$-th root of unity has degree $\phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2\cos(2\pi/n)$ has degree $\phi(n)/2$ is also abelian. Hence since $\phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
