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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
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Verified Answer
The correct answer is:
$2 \pi$
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
$$
\begin{aligned}
& =\frac{2 \pi}{3}+\frac{4 \pi}{3}=\frac{6 \pi}{3} \\
& =2 \pi
\end{aligned}
$$
$$
\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4} \cos 3 \theta
$$
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
$$
\begin{aligned}
& =\frac{2 \pi}{3}+\frac{4 \pi}{3}=\frac{6 \pi}{3} \\
& =2 \pi
\end{aligned}
$$
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