Finding the roots of any cubic with trigonometric roots. 
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?


I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\pi}7\tag{1}$$$$x^3-x^2-9x+1=0\\x_1=4\cos\frac {2\pi}7,x_2=4\cos\frac {4\pi}7+1,x_3=4\cos\frac {8\pi}7+1\tag{2}$$$$x^3+x^2-6x-7=0\\x_1=2\left(\cos\frac {4\pi}{19}+\cos\frac {6\pi}{19}+\cos\frac {10\pi}{19}\right)\\x_2=2\left(\cos\frac {2\pi}{19}+\cos\frac {14\pi}{19}+\cos\frac {16\pi}{19}\right)\\x_3=2\left(\cos\frac {8\pi}{19}+\cos\frac {12\pi}{19}+\cos\frac {20\pi}{19}\right)\tag{3}$$
With $(3)$ being a very famous relation with this problem.
Now that I know it's possible, I'm wondering if there is a simple way to find the roots of any cubic with trigonometric roots. And is it possible to use the method to find the roots of cubics such as $x^3+x^2-10x-8=0$?

For $x^3+x^2-10x-8 = 0$. 
$(\Bbb Z/31 \Bbb Z)^*/\{\pm 1\}$ is cyclic of order $15$, so it has only one subgroup $H$ of index $3$.
$H = \{\pm 1; \pm 2; \pm 4; \pm 8; \pm 15\}$ (generated by $\pm 2$)
So, the roots should be integer combinations of the quantities
$\sum_{n \in kH} 2\cos(\frac {2n\pi}{31})$ where $kH \in G/H$ are the three cosets of $H$
In fact, the roots are exactly those sums
$\sum_{n \in H} 2\cos(\frac {2n\pi}{31}) = 3.083872... \\ 
\sum_{n \in 3H} 2\cos(\frac {2n\pi}{31}) = -0.786802... \\
\sum_{n \in 9H} 2\cos(\frac {2n\pi}{31}) = -3.297071...$
However, it can't be related to an equation like $x = \sqrt {2a + \sqrt {2a +\sqrt {2a + x}}}$ because modulo $2$, the action of the Frobenius automorphism would permute the roots cyclically, while here, the Frobenius acts trivially because $\pm 2 \in H$ (or equivalently, because $2$ is a cube mod $31$).
