The general solution of $\tan θ + \tan 2 θ + \tan 3 θ = 0$ is

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$θ = nπ / 6, n \in Z$ only

$θ = nπ \pm α, n \in Z$ where $\tan α = 1 / \sqrt{2}$ only

Both 1 and 2

$θ = nπ \pm α, n \in Z$ where $\tan α = \frac{1}{\sqrt{5}}$ only

answer is B.

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Detailed Solution

From the given equation, we have
$\begin{array}{l} \tan θ + \tan 2 θ + \tan (θ + 2 θ) = 0 \\ (\tan θ + \tan 2 θ) [1 + \frac{1}{1 - \tan θtan 2 θ}] = 0 \\ \tan θ = - \tan 2 θ \Rightarrow \tan θ = \tan (- 2 θ) \\ \Rightarrow θ = nπ - 2 θ or θ = nπ / 3 \\ a n d 1 + \frac{1}{1 - \tan θtan 2 θ} = 0 \Rightarrow 1 - \tan θtan 2 θ + 1 = 0 \\ \Rightarrow \tan θtan 2 θ = 2 \Rightarrow \frac{2 \tan^{2} θ}{1 - \tan^{2} θ} = 2 \\ ∴ 1 - 2 \tan^{2} θ = 0 \\ or \tan^{2} θ = 1 / 2 = \tan^{2} α (say) \end{array}$
Therefore, $θ = nπ \pm α$ where $\tan α = 1 / \sqrt{2}$