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$x=\dfrac{2\sqrt{19}}{3} \cos \left(\dfrac{1}{3} \arccos \left(\dfrac{7}{2\sqrt{19}} \right) \right) - \dfrac{1}{3}$

$3x+1=2\sqrt{19}\cos \left(\dfrac{1}{3} \arccos \left(\dfrac{7}{2\sqrt{19}} \right) \right)$

Cube both sides and use the identity $4\cos^3 \theta=3\cos\theta+\cos3\theta$ :

$27x^3+27x^2+9x+1=152\sqrt{19}\cos^3 \left(\dfrac{1}{3} \arccos \left(\dfrac{7}{2\sqrt{19}} \right) \right)$

$27x^3+27x^2+9x+1=38\sqrt{19}\left(3\cos \left(\dfrac{1}{3} \arccos \left(\dfrac{7}{2\sqrt{19}} \right) \right)+\dfrac{7}{2\sqrt{19}}\right)$

$27x^3+27x^2+9x+1=38\sqrt{19}\left(3\left(\dfrac{3x+1}{2\sqrt{19}}\right)+\dfrac{7}{2\sqrt{19}}\right)$

Expand and simplify:

$x^3+x^2-6x-7=0$

Let $w=e^{2\pi i/19}$ be a 19th primitive root of unity. We claim that the roots of the equation above are $x_1=w+w^{18}+w^7+w^{12}+w^8+w^{11}$ , $x_2=w^2+w^{17}+w^3+w^{16}+w^5+w^{14}$ and $x_3=w^4+w^{15}+w^6+w^{13}+w^9+w^{10}$ . Let's prove it (after expanding very messy products):

$x_1+x_2+x_3=w+w^2+w^3+\cdots+w^{18}=-1$

$x_1x_2+x_1x_3+x_2x_3=6(w+w^2+w^3+\cdots+w^{18})=-6$

$x_1x_2x_3=11(w+w^2+w^3+\cdots+w^{18})+18=7$

By Vieta's formulas, the polynomial with roots $x_1,x_2$ and $x_3$ is $P(x)=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3=x^3+x^2-6x-7$ . Hence proved.

Now, prove yourself that the correct value of $x$ is $x_2$ .

Then $x=2\left(\cos\left(\dfrac{4\pi}{19}\right)+\cos\left(\dfrac{6\pi}{19}\right)+\cos\left(\dfrac{10\pi}{19}\right)\right)$

We get $a=2,b=4,c=6,d=10,e=19$ and $a+b+c+d+e=\boxed{41}$ .