Question

The sum of the solution in $[0,2\pi ]$ of the equation $\cos x\cos (\frac{\pi }{3}-x)\cos (\frac{\pi }{3}+x)=\frac{1}{4}$ is

• A. $4\pi$
• B. $\pi$
• C. $2\pi$
• D. $3\pi$

Answer 1

Answer: (A)

$4\pi$

$\cos \left(x\right)\left(\cos (\frac{\pi }{3}-x)\cos (\frac{\pi }{3}+x)\right)=\frac{1}{4}$
$\frac{\cos x}{2}\left[2\cos (\frac{\pi }{3}-x)\cos (\frac{\pi }{3}+x)\right]=\frac{1}{4}$
$\cos \left(x\right)(\cos \frac{2\pi }{3}+\cos (2x\left)\right)=\frac{1}{2}$
$\cos x(\frac{-1}{2}+2{\cos }^{2}x-1)=\frac{1}{2}$
$\cos \left(x\right)(2{\cos }^{2}x-\frac{3}{2})=\frac{1}{2}$

$2{\cos }^{3}x-\frac{3\cos x}{2}=\frac{1}{2}$
$4{\cos }^{3}x-3\cos x-1=0$
$\cos 3x-1=0$
$\cos 3x=1$
$3x=0,2\pi ,4\pi ,6\pi$
$x=0,\frac{2\pi }{3},\frac{4\pi }{3},2\pi$
Hence $0+\frac{2\pi }{3}+\frac{4\pi }{3}+2\pi$
$=\frac{6\pi }{3}+2\pi$
$=2\pi +2\pi$
$=4\pi$.