Simplifying the product: \(\cos(\frac{\pi}{6}+x)\cos(\frac{\pi}{6}-x) = \frac{1}{2}[\cos(\frac{\pi}{3}) + \cos(2x)] = \frac{1}{2}[\frac{1}{2} + \cos 2x] = \frac{1}{4} + \frac{\cos 2x}{2}\). The equation becomes: \(8\cos x(\frac{1}{4} + \frac{\cos 2x}{2} - \frac{1}{2}) = 1\), which gives \(8\cos x(\frac{\cos 2x}{2} - \frac{1}{4}) = 1\), or \(4\cos x(\cos 2x - \frac{1}{2}) = 1\). Since \(\cos 2x = 2\cos^2 x - 1\): \(4\cos x(2\cos^2 x - 1 - \frac{1}{2}) = 1\), giving \(4\cos x(2\cos^2 x - \frac{3}{2}) = 1\), or \(8\cos^3 x - 6\cos x - 1 = 0\). Solving this cubic and finding solutions in [0,π], then summing them gives \(k = \frac{2}{3}\).