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

• A. $\frac{8}{9}$
• B. $\frac{20}{9}$
• C. $\frac{2}{3}$
• D. $\frac{13}{9}$

Answer 1

The correct option is D $\frac{13}{9}$

$8\cos x[\cos (\frac{\pi }{6}+x)\cos (\frac{\pi }{6}-x)-\frac{1}{2}]=1$

using $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, we get

$8\cos x[\frac{\cos \left(\frac{\pi }{3}\right)\cos \left(2x\right)}{2}-\frac{1}{2}]=1$

$\therefore 4\cos x[\frac{1}{2}+\cos (2x)-1]=1$

$\therefore 4\cos x\left[\cos \right(2x)-\frac{1}{2}]=1$

$\therefore 4\cos x\cos \left(2x\right)-2\cos x=1$

using $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, we get

$2(\cos 3x+\cos x)-2\cos x=1$

$\therefore \cos 3x=\frac{1}{2}$

$\therefore 3x=2n\pi \pm \frac{\pi }{3}$

$\therefore x=2n\frac{\pi }{3}\pm \frac{\pi }{9}$

Solutions in $[0,\pi ]$ are $\frac{\pi }{9},\frac{2\pi }{3}-\frac{\pi }{9},\frac{2\pi }{3}+\frac{\pi }{9}$

Hence, their sum $=$ $\frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}=\frac{13\pi }{9}$

$\therefore k=\frac{13}{9}$

This is the required solution.