$\begin{aligned} X & = \cos \frac \pi 7 {\color{#3D99F6} - \cos \frac {2\pi}7} + \cos \frac {3\pi}7 & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = \cos \frac \pi 7 {\color{#3D99F6} + \cos \frac {5\pi}7} + \cos \frac {3\pi}7 & \small \color{#3D99F6} \text{and }\sum_{k=1}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi\right) = \frac 12 \text{ (see note)} \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

Note: Proof of $\displaystyle \sum_{k=1}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi\right) = \frac 12$ .

Proof for this case

$\begin{aligned} X & = {\color{#3D99F6}\cos \frac \pi 7} - \cos \frac {2\pi}7 {\color{#3D99F6} + \cos \frac {3\pi}7} & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = {\color{#3D99F6}- \cos \frac {6\pi} 7} - \cos \frac {2\pi}7 {\color{#3D99F6}- \cos \frac {4\pi}7} & \small \color{#3D99F6} \text{Let } \omega = e^{\frac {2\pi}7i} \text{, the 7th root of unity.} \\ & = - \frac {\omega^3+\color{#3D99F6}\omega^{-3}}2 - \frac {\omega+\color{#3D99F6}\omega^{-1}}2 - \frac {\omega^2+\color{#3D99F6}\omega^{-2}}2 & \small \color{#3D99F6} \text{Since }\omega^7 = 1 \\ & = - \frac 12 \left( \omega^3+ {\color{#3D99F6}\omega^4} + \omega+{\color{#3D99F6}\omega^6} + \omega^2+{\color{#3D99F6}\omega^5} \right) & \small \color{#3D99F6} \text{and } \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0 \\ & = - \frac 12 (-1) = \frac 12 = \boxed{0.5} \end{aligned}$

My method is tedious, so more nice methods are welcomed!

Check the root of $x^{7}=1,x∈C$ ①. According to Fundamental theorem of algebra , there are seven roots $x_{0},x_{1},...,x_{6}$ . $\large{\boxed{i}}$ is imaginary unit. $\begin{aligned} x^{7}=1=\cos0+i\sin0 \\ \implies x=\cos\frac{2k\pi}{7}+i\sin\frac{2k\pi}{7}②, (k=0,1,2,...,6) \\ \ By ①,\implies x^{7}-1=0,By Vieta's theorem,\sum_{i=0}^6 x_{i}=-\frac{0}{1}=0 \\ \ Back to ② \\ \implies \sum_{k=0}^6 (\cos\frac{2k\pi}{7}+i\sin\frac{2k\pi}{7})=0 \\ \implies \sum_{k=0}^6 \cos\frac{2k\pi}{7}=0 \\ \implies 0=1+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+...+\cos\frac{12\pi}{7} \\ \implies 0=1+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}-\cos\frac{\pi}{7}-\cos\frac{\pi}{7}-\cos\frac{3\pi}{7}+\cos\frac{2\pi}{7} \\ \implies \cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2} \end{aligned}$