Question

Find the sum of all solutions of $\cos x\cos (x+\frac{\pi }{3})\cos (\frac{\pi }{3}-x)=\frac{1}{4},x\in [0,6\pi ].=m\pi$.Find $m$

Answer 1

Given-

$\cos x\cos (x+\frac{\pi }{3})\cos (\frac{\pi }{3}-x)=\frac{1}{4}$

can be written as

$4\cos x\cos (x+\frac{\pi }{3})\cos (\frac{\pi }{3}-x)=1$

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

$\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)$

$\Rightarrow 2\cos x[\cos (x+\frac{\pi }{3}+\frac{\pi }{3}-x)+\cos (x+\frac{\pi }{3}-\frac{\pi }{3}+x)]=1$

$\Rightarrow 2cosx[\cos \left(\frac{2\pi }{3}\right)+\cos 2x]=1$

$\Rightarrow 2cosx[\cos (\pi -\frac{\pi }{3})+\cos 2x]=1$

$\because \cos \frac{\pi }{3}=1/2,\cos 2\theta =2{\cos }^2\theta -1$

$\Rightarrow 2cosx[\frac{-1}{2}+2{\cos }^{2}x-1]=1$

$\Rightarrow 4{\cos }^{3}x-3cosx=1$

$\because \cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$\Rightarrow \cos 3x=1=\cos \theta$

i.e. $3x=2n\pi ,n\in I$

$\Rightarrow x=\frac{2n\pi }{3},n\in I$

$x\in [0,6\pi ]$

$\Rightarrow x=[0,\frac{2\pi }{3},\frac{4\pi }{3},2\pi ,\frac{8\pi }{3},\frac{10\pi }{3},4\pi ,\frac{14\pi }{3},\frac{16\pi }{3},6\pi ]$

and their sum $=30\pi =m\pi$

$\Rightarrow m=30$