This is an exercise I got from a course of which the deadline has already passed, but I'm still confused. I already found the minimal polynomial of $\alpha := 2\cos(\frac{2\pi}{7})$ over $\mathbb{Q}$, which is $x^3 + x^2 -2x -1$.
I have no idea how to find the splitting field of that, however. My professor gave a hint that we need Galois theory for this, but I don't yet see how Galois theory can help here.
All I have so far is that If $K$ is the splitting field, $[K:\mathbb{Q}]$ divides $3!$, but it can't be $1$ or $2$, so it's either $3$ or $6$.
Have you heard of cyclotomic fields? One way to solve this problem is to consider the bigger extension $\mathbf{Q}(\zeta_7) \supset \mathbf{Q}$, where $\zeta_7 = e^{2\pi i/7}$. This extension is Galois and abelian, with Galois group isomorphic to $(\mathbf{Z}/7)^\times \cong \mathbf{Z}/6$. Note that $2\cos(2\pi /7) = \zeta_7 +\overline{\zeta}_7 = \zeta_7 + \zeta_7^6$, so that the field generated by $\alpha$ is a subextension. Since you already found the minimal polynomial, you can use Galois theory and the lattice of subgroups of $\mathbf{Z}/6$ to conclude.
Let me know if you need more context. :)