Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation 
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is 
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got 
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I tried to factorise it, got this 
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)$
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)$
 Should I now solve cubic?

$\pm 2,\pm 3,\pm 5$ are precisely twice the cubic residues $\!\pmod{19}$. Let $\omega=\exp\left(\frac{2\pi i}{19}\right)$. $\omega$ is al algebraic number over $\mathbb{Q}$ with degree $18$, hence by Galois theory both
$$ \sum_{k\in\{\pm 1,\pm 7,\pm 8\}}\omega^k,\qquad \sum_{k\in\{\pm 2,\pm 3,\pm 5\}}\omega^k $$
(related to the Kummer sums $\!\!\pmod{19}$) are algebraic numbers over $\mathbb{Q}$ with degree $3$. Indeed, the minimal polynomial of the second sum
is $z^3+z^2-6z-7$, and the second sum is exactly $2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right)$.
