If sum of all the solutions of the equation 8 x · ( ((π)/(6) + x) ·...

Given equation $8\cos x . \left(\cos \left(\dfrac{\pi}{6} + x\right) . \cos \left(\dfrac{\pi}{6} - x\right) - \dfrac{1}{2}\right) = 1$

We would use the formula

$\cos \left(A + B\right) . \cos \left(A - B\right) = \cos^{2}A - \sin^{2}B$

Now, $\cos \left(\dfrac{\pi}{6} + x\right) . \cos \left(\dfrac{\pi}{6} - x\right) = \cos^{2}\dfrac{\pi}{6} - \sin^{2}x$

$= \dfrac{3}{4} - \sin^{2}x$

$⇒ 8\cos x\left(\dfrac{3}{4} - \sin^{2}x - \dfrac{1}{2}\right) = 1$

$⇒ 8\cos x\left(\dfrac{1}{4} - \sin^{2}x\right) = 1$

$⇒ 8\cos x\left(\dfrac{1}{4} - \left(1 - \cos^{2}x\right)\right) = 1$

$⇒ 8\cos x\left(\cos^{2}x - \dfrac{3}{4}\right) = 1$

$⇒ 8\cos^{3}x - 6\cos x = 1$

$⇒ 2\left(4\cos^{3}x - 3\cos x\right) = 1$

$⇒ 2\cos 3x = 1$

$⇒ \cos 3x = \dfrac{1}{2}$

General Solution $3x = 2n\pi \pm \dfrac{\pi}{3}$

$⇒ x = \dfrac{2n\pi}{3} \pm \dfrac{\pi}{9}$

$x \in \left[0, \pi\right]$

$∴ x = \dfrac{\pi}{9}, \dfrac{2\pi}{3} - \dfrac{\pi}{9}, \dfrac{2\pi}{3} + \dfrac{\pi}{9}$

Sum = \dfrac{13\pi}{9}, k = \dfrac{13}{9}$