Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ How can one show that the number
$2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$
is a root of the equation
$\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?

Let $$f(x) = 2(\cos \frac{4\pi x}{19} + \cos \frac{6\pi x}{19} + \cos \frac{10\pi x}{19})$$
Then we claim that the following are true:
$$f(1) + f(4)^2 = 4$$
$$f(4) + f(16)^2 = 4$$
$$f(16) + f(64)^2 = 4$$
Noting that $\displaystyle f(4) \lt 0$, $\displaystyle f(16) \lt 0$ and $\displaystyle f(64) = f(1) \gt 0$, we can show the identity about $\displaystyle f(1)$.
In order to prove the above three identities,
We use the fact that if 
$\displaystyle P(z) = z^4 + z^{-4} + z^6 + z^{-6} + z^{10} + z^{-10}$
and if $\displaystyle w$ is a root of $\displaystyle z^{19} = -1$ then
$$P(w) + P^2(w^4) = 4$$
After some (slightly tedious, but easy) algebraic manipulation, the above identity involving $\displaystyle P(w)$ basically boils down to proving that
$$w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w = -1$$
which follows easily from the fact that
$$ 0 = w^{19} + 1 = (w+1)(w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w + 1)$$
Now since $\displaystyle P(z) = P(-z)$, the identity is also true of $\displaystyle w$ is a root of $\displaystyle z^{19} = 1$
Thus applying the identity with $\displaystyle P(w)$ three times to $\displaystyle c, c^4, c^{16}$, where $\displaystyle c = \cos \frac{\pi}{19} + i \sin \frac{\pi}{19}$ we get the above three identities involving $\displaystyle f(1), f(4), f(16)$ and thus we are done.
