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The general solution of $\tan \theta+\tan 2 \theta=\tan 3 \theta$ is
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The correct answer is:
$\theta=n \pi, n \in Z$ or $\theta=\frac{p \pi}{3}, p \in Z$
$\tan 3 \theta=\tan (2 \theta+\theta)=\tan \theta+\tan 2 \theta$
$\therefore \frac{\tan 2 \theta+\tan \theta}{1-\tan 2 \theta \tan \theta}=\tan \theta+\tan 2 \theta$
$\therefore 1-\tan 2 \theta \tan \theta=1 \Rightarrow \tan 2 \theta \tan \theta=0 \Rightarrow \tan \theta=0$
$\therefore \theta=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}$
$\therefore \frac{\tan 2 \theta+\tan \theta}{1-\tan 2 \theta \tan \theta}=\tan \theta+\tan 2 \theta$
$\therefore 1-\tan 2 \theta \tan \theta=1 \Rightarrow \tan 2 \theta \tan \theta=0 \Rightarrow \tan \theta=0$
$\therefore \theta=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}$
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