Question

In $\left(0,6\pi \right)$, find the number of solutions of the equation $\tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta$

Answer 1

$\Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{\sin 2\theta }{\cos 2\theta }+\frac{\sin 3\theta }{\cos 3\theta }=\frac{\sin \theta \sin 2\theta \sin 3\theta }{\cos \theta \cos 2\theta \cos 3\theta }\Rightarrow \sin \theta \cos 2\theta \cos 3\theta +\sin 2\theta \cos \theta \cos 3\theta +\sin 3\theta \cos \theta \cos 2\theta -\sin \theta \sin 2\theta \sin 3\theta \Rightarrow \cos 2\theta \{\sin \theta \cos 3\theta +\cos \theta \sin 3\theta \}+\sin 2\theta \{\cos \theta \cos 3\theta -\sin \theta \sin 3\theta \}=0\Rightarrow \cos 2\theta \sin (3\theta +\theta )+\sin 2\theta (3\theta +\theta )=0\Rightarrow \cos 2\theta \sin 4\theta +\sin 2\theta \cos 4\theta =0\Rightarrow \sin (2\theta +4\theta )=0\Rightarrow \sin \left(6\theta \right)=0\theta =\frac{n\pi }{6},nϵz$