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If $\sqrt{3} \tan 2 \theta+\sqrt{3} \tan 3 \theta+\tan 2 \theta \tan 3 \theta=1$, then find the general value of $\theta$.
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Verified Answer
$\sqrt{3} \tan 2 \theta+\sqrt{3} \tan 3 \theta+\tan 2 \theta \tan 3 \theta=1$
$$
\begin{aligned}
&\Rightarrow \frac{\tan 2 \theta+\tan 3 \theta}{1-\tan 2 \theta \tan 3 \theta}=\frac{1}{\sqrt{3}} \Rightarrow \tan 5 \theta=\tan \frac{\pi}{6} \\
&\Rightarrow 5 \theta=\mathrm{n} \pi+\frac{\pi}{6} \Rightarrow \theta=\left(\mathrm{n}+\frac{1}{6}\right) \frac{\pi}{5}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{\tan 2 \theta+\tan 3 \theta}{1-\tan 2 \theta \tan 3 \theta}=\frac{1}{\sqrt{3}} \Rightarrow \tan 5 \theta=\tan \frac{\pi}{6} \\
&\Rightarrow 5 \theta=\mathrm{n} \pi+\frac{\pi}{6} \Rightarrow \theta=\left(\mathrm{n}+\frac{1}{6}\right) \frac{\pi}{5}
\end{aligned}
$$
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