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The general value of $θ$ satisfies the equation $T a n θ T a n (120^{0} + θ) T a n (120^{0} - θ) = \frac{1}{\sqrt{3}}$ is

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$(6 n + 1) \frac{π}{18}, \forall n \in Z$

$(6 n + 1) \frac{π}{6}, \forall n \in Z$

$(3 n + 1) \frac{π}{6}, \forall n \in Z$

$(3 n + 1) \frac{π}{3}, \forall n \in Z$

answer is A.

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Detailed Solution

$\tan θ \tan (120^{0} + θ) \tan (120^{0} - θ) = \frac{1}{\sqrt{3}}$

$\Rightarrow \tan 3 θ = \frac{1}{\sqrt{3}} \Rightarrow 3 θ = n π + \frac{π}{6} \Rightarrow θ = \frac{n π}{3} + \frac{π}{18}$

$= (n + \frac{1}{6}) \frac{π}{3} = (6 n + 1) \frac{π}{18}$

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