Q100 / 160 · AP EAMCET 2014 · Mathematics

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

  • A
    $4 \pi$
  • B
    $\pi$
  • Correct answer
    C
    $2 \pi$
  • D
    $3 \pi$

Solution

We know, $$ \cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4} \cos 3 \theta $$ But it is given $$ \begin{aligned} & \cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4} \\ \therefore \quad \cos 3 \theta & =1 \\ \Rightarrow \quad 3 \theta & =2 \pi, 4 \pi \\ \Rightarrow \quad \theta & =\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{aligned} $$ $\therefore$ Sum of solutions $$ \begin{aligned} & =\frac{2 \pi}{3}+\frac{4 \pi}{3}=\frac{6 \pi}{3} \\ & =2 \pi \end{aligned} $$

Hence, the correct answer is option: 'C'.

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