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)

$8 \cos x \cdot\left(\cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}\right)=1$
$\Rightarrow 8 \cos x\left(\frac{3}{4}-\frac{1}{2}-1+\cos ^{2} x\right)=1$
$\Rightarrow 8 \cos x\left(\frac{-3+4 \cos ^{2} x}{4}\right)=1$
$\Rightarrow 2 \cos 3 x=1$
$\Rightarrow \cos 3 x=\frac{1}{2}$
$\Rightarrow 3 x=\frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3}$
$\Rightarrow x=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}$
$\Rightarrow $ Sum $=\frac{13 \pi}{9}$
$\Rightarrow k=\frac{13}{9}$