Waiting for 2016 - 2

$x=\dfrac{4\sqrt{7}}{3} \cos \left(\dfrac{1}{3} \arccos \left(\dfrac{1}{2\sqrt{7}} \right) \right) + \dfrac{1}{3}$

$3x-1=4\sqrt{7}\cos \left(\dfrac{1}{3} \arccos \left(\dfrac{1}{2\sqrt{7}} \right) \right)$

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

$27x^3-27x^2+9x-1=448\sqrt{7}\cos^3 \left(\dfrac{1}{3} \arccos \left(\dfrac{1}{2\sqrt{7}} \right) \right)$

$27x^3-27x^2+9x-1=112\sqrt{7}\left(3\cos \left(\dfrac{1}{3} \arccos \left(\dfrac{1}{2\sqrt{7}} \right) \right)+\dfrac{1}{2\sqrt{7}}\right)$

$27x^3-27x^2+9x-1=112\sqrt{7}\left(3\left(\dfrac{3x-1}{4\sqrt{7}}\right)+\dfrac{1}{2\sqrt{7}}\right)$

Expand and simplify:

$x^3-x^2-9x+1=0$

Now let $x=2y+1$ :

$(2y+1)^3-(2y+1)^2-9(2y+1)+1=0$

Expand and simplify:

$y^3+y^2-2y-1=0$

It's a well known fact that that equation has roots $y_k=2\cos\left(\dfrac{2\pi k}{7}\right)$ for $k=\{1,2,3\}$ . So the roots of the first equation are $x_k=4\cos\left(\dfrac{2\pi k}{7}\right)+1$ . Prove yourself that the value we are looking for is when $k=1$ . So $x=4\cos\left(\dfrac{2\pi}{7}\right)+1$ .

But $1=-2\left(\cos\left(\dfrac{2\pi}{7}\right)+\cos\left(\dfrac{4\pi}{7}\right)+\cos\left(\dfrac{6\pi}{7}\right)\right)$ , so $x=2\left(\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\dfrac{4\pi}{7}\right)-\cos\left(\dfrac{6\pi}{7}\right)\right)$ . Using $-\cos\theta=\cos(\pi-\theta)$ we get finally:

$x=2\left(\cos\left(\dfrac{\pi}{7}\right)+\cos\left(\dfrac{2\pi}{7}\right)+\cos\left(\dfrac{3\pi}{7}\right)\right)$

Hence $a=2,b=1,c=2,d=3,e=7$ and $a+b+c+d+e=\boxed{15}$ .