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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.
HINT:
$$\tan\theta = 2\sin\theta \iff \sin \theta = 2 \sin \theta \cos \theta \iff \sin\theta(1 - 2\cos\theta)=0$$
ADDED: I am rewriting this in a form where you can "read off" solutions. $$ab = 0\; \iff \;a = 0 \;\text{ or }\; b = 0$$ So in the case at hand, we have that $$\sin\theta(1 - 2\cos\theta)=0 $$ $$\iff\sin \theta = 0,\;\text{ or } \; 1 - 2\cos \theta = 0 $$ $$\iff \sin \theta = 0 \;\text{ or }\; \cos\theta = \frac 12$$
Note that so long as $\theta$ is such that $\cos\theta\ne 0,$ we have $$\tan\theta=\frac{\sin\theta}{\cos\theta},$$ so your problem reduces to finding all $\theta$ such that $\cos\theta\ne 0$ and $$\frac{\sin\theta}{\cos\theta}=2\sin\theta.\tag{$\star$}$$ Note that we can rewrite $(\star)$ as $$\left(\frac1{\cos\theta}-2\right)\sin\theta=0.\tag{$\heartsuit$}$$ Can you take it from there?
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.
The equation you wish to solve can be rewritten $$\frac{\sin t}{\cos t}=2\sin\theta.$$
If $\cos t=0$, then it is meaningless, so we may have to watch out for such "false solutions" later.
If $\sin t=0$, then the equation certainly holds.
Suppose $\sin t\ne 0$.
Then we can divide through by that, getting $\frac 1 {\cos t}=2$, and rearranging find $\cos t=\frac12$ (so we had no need to worry about false solutions).
Thus the solutions are values of $t$ for which $\sin t=0$ or $\cos t=1/2$.
When $\sin\alpha=\sin\beta$, you either have $$ \beta=\alpha+2k\pi \quad(k\in\mathbb{Z}) $$ or $$ \beta=\pi-\alpha+2k\pi \quad(k\in\mathbb{Z}). $$ In your case, this becomes $$ 2\theta=\theta+2k\pi \quad(k\in\mathbb{Z}) $$ that is, $\theta=2k\pi$, or $$ 2\theta=\pi-\theta+2k\pi \quad(k\in\mathbb{Z}) $$ that is $$ \theta=\frac{\pi}{3}+\frac{2}{3}k\pi \quad(k\in\mathbb{Z}) $$ For the original equation to be satisfied, we need to ensure that, given $k$, $$ \frac{\pi}{3}+\frac{2}{3}k\pi\ne\frac{\pi}{2}+h\pi $$ for any integer $h$. The equality would mean $2+4k=3+6h$, that is $6h-4k=1$, which has no solution. Therefore no $k$ has to be excluded.
In this particular case, rewriting the equation as $\sin\theta(1-2\cos\theta)=0$ is surely better. In other cases, such as $\sin\theta=\sin4\theta$, this may be handier.
