<s>[INST] Solve the equation $\tan \theta = 2\sin \theta$. 
Solve the equation $\tan \theta  = 2\sin \theta$. 

What I did was rewrite it to the form $$\sin \theta  = 2 \sin \theta \cos \theta$$ You'll get $$\sin \theta = \sin\ 2 \theta.$$
How am I supposed to solve this when I have $\sin$ on both sides? My main problem with these types of 'solve' equations are that I don't know which forms I should rewrite them too. Usually just to the form $\sin \theta = n$, but I wonder if having a $\sin$ on both sides can result in an answer too. 
 [/INST] We have the equality $$\sin x - \sin y = 2\sin \frac{x - y}{2} \cos\frac{x + y}{2}.$$ Hence, $\sin x = \sin y$ if and only if either $\sin \frac{x-y}{2} = 0$ or $\cos \frac{x+y}{2} = 0$. 
In particular, for the solutions of $\sin\theta = \sin(2\theta)$, substitute $x=\theta$, $y=2\theta$. We need $\sin(\theta/2) = 0$ or $\cos(3\theta/2) = 0$. Now, $$\sin(\theta/2) = 0 \iff \exists n\in\mathbb{Z},\,\theta/2 = \pi n \iff \exists n,\,\theta = 2\pi n,$$ and $$\cos(3\theta/2) = 0 \iff \exists n,\,3\theta/2 = \frac{\pi}{2}+\pi n \iff \exists n,\,\theta = \frac{\pi}{3}+\frac{2\pi n}{3}.$$ The solution set is thus
$$\left\{2\pi n:n\in\mathbb{Z}\right\}\cup\left\{\frac{\pi}{3}+\frac{2\pi n}{3}:n\in\mathbb{Z}\right\}.$$
Finally, we verify that $\frac{\pi}{2}+\mathbb{Z}\pi$ is not a subset of our solution set, hence the $\tan$ is well-defined.
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