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$\begin{aligned} X & = 8\cos \frac {2\pi}7 \left(2 \cos^2 \frac {2\pi}7 + \cos \frac {2\pi}7 - 1\right) + 1 \\ & = 16 \cos^3 \frac {2\pi}7 + 8 \cos^2 \frac {2\pi}7 - 8 \cos \frac {2\pi}7 + 1 \\ & = 4\left(4\cos^3 \frac {2\pi}7 - 3\cos \frac {2\pi}7 \right) + 4\left(2\cos^2 \frac {2\pi}7 - 1\right) + 4 \cos \frac {2\pi}7 + 5 \\ & = 4\left(\cos \frac {6\pi}7 + \cos \frac {4\pi}7 + \cos \frac {2\pi}7 \right) + 5 & \small \color{#3D99F6} \text{Note that } \cos (\pi - x) = - \cos x \\ & = - 4\left({\color{#3D99F6}\cos \frac \pi 7 + \cos \frac {3\pi}7 + \cos \frac {5\pi}7} \right) + 5 & \small \color{#3D99F6} \text{See note: }\sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 \\ & = - 4\left({\color{#3D99F6}\frac 12} \right) + 5 \\ & = \boxed{3} \end{aligned}$

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Note:

$\small \begin{aligned} S & = \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) \\ & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k\pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left(\frac {1-e^{\frac {2n\pi i}{2n+1}}}{1-e^{\frac {2\pi i}{2n+1}}}\right) \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}}-e^{\pi i}}{1-e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}} + 1}{\left(1+e^{\frac {\pi i}{2n+1}} \right) \left(1-e^{\frac {\pi i}{2n+1}} \right)} \right \} \\ & = \Re \left \{ \frac 1{1-e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac 1{1-\cos \frac {\pi i}{2n+1} - i\sin \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{\left(1-\cos \frac {\pi i}{2n+1}\right)^2 + \sin^2 \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{2-2\cos \frac {\pi i}{2n+1}} \right \} \\ & = \frac 12 \end{aligned}$

Let $\cos(7x)=\cos^7(x)-21\cos^5(x)\sin^2(x)+35\cos^3(x)\sin^4(x)-7\cos(x)\sin^6(x)$

As we can see that for $x=\frac{2\pi}{7}$ , LHS is $1$ So RHS must be (lets write $\cos(x)=t$

$1=t^7-21t^5(1-t^2)+35t^3(1-t^2)^2-7t(1-t^2)^3$

Roots of this equation are $\cos(\frac{2n\pi}{7})$ where $n$ is $0,1,2,3,4,5,6$ . Also $n=6$ gives same result as $n=1$ (As $\cos(2\pi-x)=\cos(x)$ And similarly for n=5 and n=4. This means that after removing $t=1$ which is $n=0$ , All the other roots are repeated twice. This means that we shall get a cubic squared.

Then after expanding all the terms in the centre equation, we get something like

$0=(t-1)(8t^3+4t^2-4t-1)^2$

As $x=\frac{2\pi}{7}$ is a root, and it does not satisfy $t-1=0$ It must satisfy the cubic equation.

Thus , for $x=\frac{2\pi}{7}$ we have $4t(2t^2+t-1)-1=0$

$8t(2t^2+t-1)+1=3$

Hence proved.

For finding expansion of $\cos(7x)$ and for finding the roots of the equation, I have used De Moivre's Theorem.

From this solution, we can see that $x=\frac{2\pi}{7}$ , $x=\frac{4\pi}{7}$ $x=\frac{6\pi}{7}$ are the 3 values for which the expression has the same answer i.e 3

pi/7 = x

8cos(2x)[cos4x+cos2x]+1

8cos(2x)[2cos3xcosx]+1

16cosxcos2xcos3x+1

note that cos3x = -cos4x

-16cosxcos2xcos4x+1

Now use the formula of product cosine series wherein angles are in GP

to get the product = -1/8

so answer is 2+1 = 3

If $\zeta =e^{\frac{2\pi i}{7}}$ then $\zeta^7 = 1$ and $\zeta \neq 1$ , so that $\zeta^6 + \zeta^5 + \zeta^4 + \zeta^3 + \zeta^2 + \zeta +1 = 0$ . Hence $\begin{aligned} 0 & = \zeta^3 + \zeta^2 + \zeta + 1 + \zeta^{-1} + \zeta^{-2} + \zeta^{-3} \; = (\zeta + \zeta^{-1})^3 + (\zeta+\zeta^{-1})^2 - 2(\zeta+\zeta^{-1}) - 1 \\ & = 8c^3 + 4c^2 - 2c - 1 \; = \; 4c(2c^2 + c - 1) - 1 \end{aligned}$ where $c = \cos\tfrac{2\pi}{7}$ . Thus $8c(2c^2 + c - 1) + 1 \,=\, 2\times1+1 = \boxed{3}$ .

2cos2π/7(cos4π/7+ cos2π/7) + 1 Used formula for cos2A Now use cosA+cosB = 2cos(A+B)/2 cos(A-B)/2 We get 16cos2π/7cos3π/7cosπ/7 +1 We get -16cosπ/7cos2π/7cos4π/7+1 Dividing and multiply by sinπ/7 We get on further solving using property of sin2A -2sin8π/7 ÷ sinπ/7 +1 We get 2+1 = 3