Question

The sum of all the solution of the equation $cosx.cos\left(\right(\pi /3)+x).cos\left(\right(\pi /3)-x)=1/4,xϵ[0,6\pi ]$ is-

• A. $15\pi$
• B. $30\pi$
• C. $110\pi /3$
• D. None of these

Answer 1

The correct option is B $30\pi$

$\cos x\cdot \cos x\left(\right(\frac{\pi }{3}+x)\cdot \cos ((\frac{\pi }{3}-x)=\frac{1}{4},x\in [0,6\pi ]$

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

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

or, $2\cos x[\cos \frac{2\pi }{3}+\cos 2x]=1$

or, $2\cos x[\frac{-1}{2}+\cos 2x]=1$

or, $-\cos x+\cos 3x+\cos x=1$

or, $\cos 3x=1=\cos 0$

or, $3x=2n\pi \pm 0$ where $n=0,1,2,\mathrm{3.......}$

or, $x=2n\frac{\pi }{3}$

$x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{6\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3},\frac{12\pi }{3},\frac{14\pi }{3},\frac{16\pi }{3},\frac{18\pi }{3}$

Sum of all values = $\frac{2\pi [1+2+3+4+5+6+7+8+9]}{3}$

$=\frac{2\pi \times 45}{3}$

$=30\pi$