Question

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

• A. $15\pi$
• B. $30\pi$
• C. $\frac{110\pi }{3}$
• D. none of these

Answer 1

The correct option is B $30\pi$

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

$\Rightarrow cosx.(cos\frac{\pi }{3}.cosx+sin\frac{\pi }{3}.sinx).(cos\frac{\pi }{3}.cosx-sin\frac{\pi }{3}.sinx)=\frac{1}{4}$

$\Rightarrow cosx.(\frac{1}{4}co{s}^{2}x-\frac{3}{4}si{n}^{2}x)=\frac{1}{4}$

$\Rightarrow cosx.(\frac{1}{4}co{s}^{2}x-\frac{3}{4}(1-co{s}^{2}x\left)\right)=\frac{1}{4}$

$\Rightarrow cosx.(\frac{1}{4}co{s}^{2}x+\frac{3}{4}co{s}^{2}x-\frac{3}{4})=\frac{1}{4}$

$\Rightarrow cosx.(co{s}^{2}x-\frac{3}{4})=\frac{1}{4}$

$\Rightarrow cosx.(4co{s}^{2}x-3)=1\Rightarrow 4co{s}^{3}x-3cosx=1$

$\Rightarrow cos3x=1$

Therefore general solution is,

$3x=2n\pi \pm 0\Rightarrow x=\frac{2n\pi }{3}$, where n is any integer

thus solutions in the given interval are,

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

$sum=0+\frac{2\pi }{3}+\frac{4\pi }{3}+2\pi +\frac{8\pi }{3}+\frac{10\pi }{3}+4\pi +\frac{14\pi }{3}+\frac{16\pi }{3}+6\pi$

$sum=12\pi +\frac{54\pi }{3}$

$sum=\frac{90\pi }{3}$

$sum=30\pi$

Sum of these solution is $30\pi$
Hence,option $B$ is correct.