$\frac{\pi}{7}$

We know that $cosx=-cos(\pi-x)$ and we also know that $sinx=sin(\pi-x)$ . So we can write $cosx+cos3x+cos5x=\dfrac{sin6x}{2sinx}.....(1)$ Substituting $x=\dfrac{\pi}{7}$ ,we get $cos\frac{\pi}{7}+cos\frac{3\pi}{7}+cos\frac{5\pi}{7}=\frac{sin6\frac{\pi}{7}}{2sin\frac{\pi}{7}}.....(2)$ ..and since $cosx=-cos(\pi-x)$ ,so $cos\frac{5\pi}{7}=-cos\bigg(\pi-\frac{5\pi}{7}\bigg)=-cos\bigg(\frac{2\pi}{7}\bigg)$ .. Similarly , we convert $sin\frac{6\pi}{7}$ by using the identity of $sinx$ as mentioned above. We get $sin\frac{6\pi}{7}=sin\frac{\pi}{7}$ . Now substituting it in equation $(1)$ ,we get $cos\dfrac{\pi}{7}-cos\dfrac{2\pi}{7}+cos\dfrac{3\pi}{7}=\frac{1}{2}$ and that is equal to $\boxed{0.5}$

Note that

$\sin(\pi-\alpha)=\sin\alpha,~~\sin(\pi+\alpha)=-\sin\alpha, ~~\text{and}~\cos(\pi+\alpha)=-\cos\alpha$

$\cos\dfrac{\pi}{7}-\cos\dfrac{2\pi}{7}+\cos\dfrac{3\pi}{7}=\cos\dfrac{\pi}{7}+\cos\dfrac{9\pi}{7}+\cos\dfrac{3\pi}{7}$

$A=2\cos\dfrac{\pi}{7}\sin\dfrac{\pi}{7}=\sin\dfrac{2\pi}{7}\\B=2\cos\dfrac{9\pi}{7}\sin\dfrac{\pi}{7}=\sin\dfrac{10\pi}{7}-\sin\dfrac{8\pi}{7}=-\dfrac{3\pi}{7}+\sin\dfrac{\pi}{7}\\C=2\cos\dfrac{3\pi}{7}\sin\dfrac{\pi}{7}=\sin\dfrac{4\pi}{7}-\sin\dfrac{2\pi}{7}=\sin\dfrac{3\pi}{7}-\sin\dfrac{2\pi}{7}\\A+B+C=2\sin\dfrac{\pi}{7}\left(\cos\dfrac{\pi}{7}+\cos\dfrac{9\pi}{7}+\cos\dfrac{3\pi}{7}\right)=\sin\dfrac{\pi}{7}\\\cos\dfrac{\pi}{7}+\cos\dfrac{9\pi}{7}+\cos\dfrac{3\pi}{7}=\dfrac{1}{2}$