Question

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

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

Answer 1

Answer: (B)

$\frac{13}{9}$

Answer 2

$8 \cos x \cdot\left(\cos \frac{\pi}{6}-\sin x-\frac{1}{2}\right)=1$

$\Rightarrow 8 \cos x\frac{\left(\cos\frac\pi2+\cos2x - \frac12\right)}{2}=1$

$4\cos x\cos 2x - 2\cos x =1$

$2(\cos 3x+ \cos3x)= 2\cos x =1$

$\therefore \cos 3x = \frac12$

$\therefore 3x=2n\pi+\frac\pi3$

$\therefore 2n \frac\pi3+\frac\pi9$

$\therefore [0,\pi] are\frac\pi9,2\frac\pi3-\frac\pi9,2\frac\pi3+\frac\pi9$

their sum is

$\frac\pi9+8\frac\pi9+7\frac\pi9=13\frac\pi9$

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