Let $$x=\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{9\pi}{13}$$ and $$y=\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{11\pi}{13}$$ Then by %%HREF%%3869%%, we find $$x+y=\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}=\frac{1}{2}$$ and \begin{align} xy&=\Big(\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\frac{9\pi}{13}\Big)\Big(\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\frac{11\pi}{13}\Big)\\ &=-\frac{3}{2}\cdot\Big(\cos\frac{\pi}{13}-\cos\frac{2\pi}{13}+\cos\frac{3\pi}{13}-\cos\frac{4\pi}{13}+\cos\frac{5\pi}{13}-\cos\frac{6\pi}{13}\Big)\\ &=-\frac{3}{2}\cdot\Big(\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}\Big)\\ &=-\frac{3}{4} \end{align} Therefore by Vieta's theorem, $x$ and $y$ are the two roots of the following equation $$u^2 -\frac{1}{2}u-\frac{3}{4}=0$$ Because apparently, $x> 0$, we find $x=\boxed{\frac{1+\sqrt{13}}{4}}$.