Question

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

Answer 1

Given:$8\cos x(\cos (\frac{\pi }{6}+x)\cos (\frac{\pi }{6}-x)-\frac{1}{2})=1$
$\Rightarrow 8\cos x({\cos }^2\frac{\pi }{6}-{\sin }^{2}x-\frac{1}{2})=1$
$\Rightarrow 6\cos x-8\cos x{\sin }^{2}x-4\cos x=1$
$\Rightarrow 2\cos x-8\cos x(1-{\cos }^{2}x)=1$
$\Rightarrow 4{\cos }^{3}x-3\cos x=\frac{1}{2}$
$\Rightarrow \cos 3x=\frac{1}{2}$
$\Rightarrow 3x=2n\pi \pm \frac{\pi }{3}$
$\Rightarrow x=\frac{2n\pi }{3}\pm \frac{\pi }{9}$
$\Rightarrow x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$
Now, $k\pi =\frac{13\pi }{9}$
$\therefore k=\frac{13}{9}$
Hence, The correct answer is option C.