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

Note that $\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B$

We have, $8\cos x\left[\cos (\pi /6+x)\cos (\pi /6-x)-1/2\right]=1$

$\begin{array}{l}\Rightarrow 8\cos x\left\{{\cos }^2\frac{\pi }{6}-{\sin }^{2}x-1/2\right\}=1\\ \Rightarrow 8\cos x\left(\frac{3}{4}-1+{\cos }^{2}x-\frac{1}{2}\right)=1\end{array}$

$\begin{array}{l}\Rightarrow 8\cos x\left({\cos }^{2}x-3/4\right)=1\Rightarrow 8\cos x\frac{\left(4{\cos }^{2}x-3\right)}{4}=1\\ \Rightarrow \left(4{\cos }^{3}x-3\cos x\right)=1/2\Rightarrow \cos 3x=1/2\end{array}$

As $x\in [0,\pi ]=3x\in [0,3\pi ]$ which gives

$3x=\pi /3,5\pi /3,7\pi /3\Rightarrow x=\pi /9,5\pi /9,7\pi /9$

which gives sum  of all values of $=\frac{13\pi }{9}\therefore k=\frac{13}{9}$