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$
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.
This solution is basically the stupidest possible way to solve the problem. I'm just posting it in case the direct calculation here sheds any light on the problem, it probably doesn't though. Especially frustrating is this does not seem to show any connection between the discriminant $19^2$ and the cyclotomic field involved.
Define $e(q) = e^{2 i \pi q}$ so that $2 \cos(2 \pi q) = e(q) + e(-q)$ and $x = 2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right) = e(2/19) + e(3/19) + e(5/19) + \text{mi}$. "mi" means mirror image, I'm going to omit these terms because there are a lot of terms involved. Since $x$ is real its square and cube will be too, that means (by algebraic independence of roots of unity) we must have the mirror image there too.. this save a bit of work.
Squaring it is a simple process it just means multiplying out the bracket and using the identity $e(q)e(r) = e(q+r)$. We get
x^2 = 6 + 2 e(1/19) + 2 e(2/19) + 2 e(3/19)
+ e(4/19) + 2 e(5/19) + e(6/19)
+ 2 e(7/19) + 2 e(8/19) + e(9/19) + mi
Cubing is slightly more work but in the end we get
x^3 = 12 + 9 e(1/19) + 15 e(2/19) + 15 e(3/19)
+ 10 e(4/19) + 15 e(5/19) + 10 e(6/19)
+ 9 e(7/19) + 9 e(8/19) + 10 e(9/19) + mi
Note that I have also reduced terms, e.g. $e(11/19) = e((11-19)/19) = e(-8)$. There is another reduction we can make too, since the terms contain the complete sum $e(1/19)+e(2/19)+...+e(9/19) + \text{mi} = -1$ (which can be seen consider the points evenly spaced around the unit circle) we apply this reduction to get:
x^2 = 4 - e(4/19) - e(6/19) - e(9/19) + mi
x^3 = 3 + 6 e(2/19) + 6 e(3/19) + e(4/19)
+ 6 e(5/19) + e(6/19) + e(9/19) + mi
It is now very easy to see that this satisfies the polynomial $x^3+x^2-6x-7$ and in fact you could have figured out that polynomial from this method.
What I would really like to do is start with the polynomial and produce an exponential series solving it..
Sorry if this is a little off-topic here, but the question where it was topical was marked as a duplicate of this one.
Well, $2$ is a primitive root $\pmod{19}$, so we can group the nonzero residues $2^n$ according to the congruence class of $n\pmod{18}$.
$$\begin{array}{ccccccc}
r_0&\color{red}{1}&\color{red}{8}&\color{red}{7}&\color{red}{18}&\color{red}{11}&\color{red}{12}\\
r_1&\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
r_2&\color{blue}{4}&\color{blue}{13}&\color{blue}{9}&\color{blue}{15}&\color{blue}{6}&\color{blue}{10}\\
\end{array}$$
From the problem statement, clearly we are interested in $r_1$. To work out the products of $r_0$, $r_1$,and $r_2$ we can write out tables were we add all the elements $\pmod{19}$. The table for $r_0\times r_1$:
$$\begin{array}{ccccccc}
&\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{red}{1}&\color{green}{3}&\color{green}{17}&\color{blue}{15}&\color{red}{18}&\color{blue}{4}&\color{blue}{6}\\
\color{red}{8}&\color{blue}{10}&\color{green}{5}&\color{green}{3}&\color{blue}{6}&\color{red}{11}&\color{blue}{13}\\
\color{red}{7}&\color{blue}{9}&\color{blue}{4}&\color{green}{2}&\color{green}{5}&\color{blue}{10}&\color{red}{12}\\
\color{red}{18}&\color{red}{1}&\color{blue}{15}&\color{blue}{13}&\color{green}{16}&\color{green}{2}&\color{blue}{4}\\
\color{red}{11}&\color{blue}{13}&\color{red}{8}&\color{blue}{6}&\color{blue}{9}&\color{green}{14}&\color{green}{16}\\
\color{red}{12}&\color{green}{14}&\color{blue}{9}&\color{red}{7}&\color{blue}{10}&\color{blue}{15}&\color{green}{17}\\
\end{array}$$
From the above we can see that $r_0\times r_1=r_0+2r_1+3r_2$. The table for $r_1\times r_1$:
$$\begin{array}{ccccccc}
&\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{green}{2}&\color{blue}{4}&\color{red}{18}&\color{green}{16}&\color{black}{0}&\color{green}{5}&\color{red}{7}\\
\color{green}{16}&\color{red}{18}&\color{blue}{13}&\color{red}{11}&\color{green}{14}&\color{black}{0}&\color{green}{2}\\
\color{green}{14}&\color{green}{16}&\color{red}{11}&\color{blue}{9}&\color{red}{12}&\color{green}{17}&\color{black}{0}\\
\color{green}{17}&\color{black}{0}&\color{green}{14}&\color{red}{12}&\color{blue}{15}&\color{red}{1}&\color{green}{3}\\
\color{green}{3}&\color{green}{5}&\color{black}{0}&\color{green}{17}&\color{red}{1}&\color{blue}{6}&\color{red}{8}\\
\color{green}{5}&\color{red}{7}&\color{green}{2}&\color{black}{0}&\color{green}{3}&\color{red}{8}&\color{blue}{10}\\
\end{array}$$
This says that $r_1\times r_1=6+2r_0+2r_1+r_2$. And finally the table for $r_2\times r_1$:
$$\begin{array}{ccccccc}
&\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{blue}{4}&\color{blue}{6}&\color{red}{1}&\color{red}{18}&\color{green}{2}&\color{red}{7}&\color{blue}{9}\\
\color{blue}{13}&\color{blue}{15}&\color{blue}{10}&\color{red}{8}&\color{red}{11}&\color{green}{16}&\color{red}{18}\\
\color{blue}{9}&\color{red}{11}&\color{blue}{6}&\color{blue}{4}&\color{red}{7}&\color{red}{12}&\color{green}{14}\\
\color{blue}{15}&\color{green}{17}&\color{red}{12}&\color{blue}{10}&\color{blue}{13}&\color{red}{18}&\color{red}{1}\\
\color{blue}{6}&\color{red}{8}&\color{green}{3}&\color{red}{1}&\color{blue}{4}&\color{blue}{9}&\color{red}{11}\\
\color{blue}{10}&\color{red}{12}&\color{red}{7}&\color{green}{5}&\color{red}{8}&\color{blue}{13}&\color{blue}{15}\\
\end{array}$$
And so $r_2\times r_1=3r_0+r_1+2r_2$. We also know by symmetry that $0=1+r_0+r_1+r_2$, and we can work out
$$\begin{align}r_1^3&=r_1\times(6+2r_0+2r_1+r_2)\\
&=6r_1+2(r_0+2r_1+3r_2)+2(6+2r_0+2r_1+r_2)+3r_0+r_1+2r_2\\
&=12+9r_0+15r_1+10r_2\end{align}$$
so we can set up the system
$$\begin{array}{ccccc}r_1^3&=12&+9r_0&+15r_1&+10r_2\\
r_1^2&=6&+2r_0&+2r_1&+r_2\\
r_1&=&&r_1&\\
0&=1&+r_0&+r_1&+r_2\end{array}$$
Eliminating $r_0$ and $r_2$, we find that
$$r_1^3+r_1^2-6r_1-7=0$$
Let $r_1=x-\frac13$
$$x^3-\frac{19}3x-\frac{133}{27}=0$$
Using Vieta's trigonometric solution we find that
$$r_1=-\frac13+\frac23\sqrt{19}\cos\left(\frac13\cos^{-1}\left(\frac7{2\sqrt{19}}\right)+\frac{2\pi k}3\right)$$
Comparing numerical values, $k=0$ corresponds to $r_1$, $k=1$ corresponds to $r_2$, and $k=2$ corresponds to $r_0$.