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)

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