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The sum of all the solution of the equation $\cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)=\frac{1}{4}, \theta \in[0,6 \pi]$
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Verified Answer
The correct answer is:
$30 \pi$
$30 \pi$
$$
\begin{aligned}
& 2 \cos \theta\left[\cos 120^{\circ}+\cos 2 \theta\right]=1 \\
\Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\
\Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\
\Rightarrow & \cos 3 \theta=1=\cos 0 \\
\Rightarrow & 3 \theta=2 n \pi \text { or } \theta=\frac{2 n \pi}{3}, n \in z
\end{aligned}
$$
Given the values so that $2 n$ does not exceed 18 .
Hence, the sum $=\frac{2 \pi}{3} \sum_1^9 n=\frac{2 \pi}{3} \times \frac{9(9+1)}{2}=30 \pi$
\begin{aligned}
& 2 \cos \theta\left[\cos 120^{\circ}+\cos 2 \theta\right]=1 \\
\Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\
\Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\
\Rightarrow & \cos 3 \theta=1=\cos 0 \\
\Rightarrow & 3 \theta=2 n \pi \text { or } \theta=\frac{2 n \pi}{3}, n \in z
\end{aligned}
$$
Given the values so that $2 n$ does not exceed 18 .
Hence, the sum $=\frac{2 \pi}{3} \sum_1^9 n=\frac{2 \pi}{3} \times \frac{9(9+1)}{2}=30 \pi$
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