As we know, $\cos \left( {x + y} \right)\cos \left( {x - y} \right) = {\cos ^2}x - {\sin ^2}y$ $\therefore$ ${\mathop{\rm cos}\nolimits} \left( {{\pi \over 6} + x} \right)cos\left( {{\pi \over 6} - x} \right) = {\cos ^2}\left( {{\pi \over 6}} \right) - {\sin ^2}x$ Given, $8\cos x\left( {\cos \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right) - {1 \over 2}} \right) = 1$ $\Rightarrow 8\cos x\left( {{{\cos }^2}{\pi \over 6} - {{\sin }^2}x - {1 \over 2}} \right) = 1$ $\Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - {{\sin }^2}x} \right) = 1$ $\Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - 1 + {{\cos }^2}x} \right) = 1$ $\Rightarrow 8\cos x\left( {{{\cos }^2}x - {3 \over 4}} \right) = 1$ $\Rightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1$ (Taking $4\cos x$ inside the bracket). We know, $4{\cos ^3}x - 3\cos x = \cos 3x$ $\Rightarrow 2\cos 3x = 1$ $\Rightarrow \cos 3x = {1 \over 2}$ $\therefore$ $\,\,\,$ $3x = 2n\pi \pm {\pi \over 3}$ So, $x = {{2n\pi } \over 3} \pm {\pi \over 9}$ At $n=0,$ $x = + {\pi \over 9}$ as $x \in \left[ {0,\pi } \right]$ At $n=1,$ $x = {{2\pi } \over 3} + {\pi \over 9}$ $\therefore$ $\,\,\,$ $x = {{2\pi } \over 3} + {\pi \over 9}$ and $x = {{2\pi } \over 3} - {\pi \over 9}$ At $n = 2,$ $x = {{4\pi } \over 3} \pm {\pi \over 9}$ Which does not belongs to $\left[ {0,\pi } \right]$ So, possible values of $x$ is $= {\pi \over 9},{{2\pi } \over 3} + {\pi \over 9},{{2\pi } \over 3} - {\pi \over 3}$ Sum of the solutions $= {\pi \over 9} + {{2\pi } \over 3} + {\pi \over 9} + {{2\pi } \over 3} - {\pi \over 9}$ v $= {{4\pi } \over 3} + {\pi \over 9}$ $= {{13\,\pi } \over 9}$ According to the question, $k\pi = {{13\pi } \over 9}$ $\therefore\,\,\,$ $k = {{13} \over 9}$