If sum of all the solutions of the equation  $8\cos x\left\{\cos \left(\frac{\pi }{6}+x\right)\cos \left(\frac{\pi }{6}-x\right)-\frac{1}{2}\right\}=1$ in $[0,\pi ]$is $k\pi$ Then ,$k$is equal to 

We have, 

$\begin{array}{l}8\cos x\left\{\cos \left(\frac{\pi }{6}+x\right)\cos \left(\frac{\pi }{6}-x\right)-\frac{1}{2}\right\}=1\\ \Rightarrow 8\cos x\left({\cos }^2\frac{\pi }{6}-{\sin }^{2}x-\frac{1}{2}\right)=1\\ \Rightarrow 8\cos x\left(\frac{1}{4}-{\sin }^{2}x\right)=1\end{array}$ 

$\begin{array}{l}\Rightarrow 8\cos x\left({\cos }^{2}x-\frac{3}{4}\right)=1\\ \Rightarrow 2\left(4{\cos }^{3}x-3\cos x\right)=1\\ \Rightarrow \cos 3x=\frac{1}{2}\Rightarrow \cos 3x=\cos \frac{\pi }{3}\Rightarrow 3x=2n\pi \pm \frac{\pi }{3}\\ \Rightarrow x=\frac{2}{3}n\pi \pm \frac{\pi }{9},n\in Z\end{array}$

$\Rightarrow x=\frac{2}{3}n\pi \pm \frac{\pi }{9},n\in Z$

$\Rightarrow x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}[\because x\in [0,\pi \left]\right]$

$\therefore k\pi =\frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}\Rightarrow k=\frac{13}{9}$