Why do we lose solutions when we divide $2\sin\theta\cos\theta=\sin\theta$ by $\sin\theta$?

Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by $0$ is not valid.

So going from $2 \sin \theta \cos \theta = \sin \theta$ to $2 \cos\theta = 1$ is only valid when $\sin\theta \ne 0$.

In general, all this means is that you need to check the $\sin \theta = 0$ case separately. For example, going from $2 \sin\theta \cos\theta = 1$ to $2 \cos\theta = \frac1{\sin\theta}$ is also only valid when $\sin\theta \ne 0$, but you don't lose any solutions by doing this, because values of $\theta$ for which $\sin\theta=0$ weren't solutions to begin with.

But in this particular case, when $\sin\theta = 0$, the equation $2\sin\theta \cos\theta = \sin\theta$ is satisfied, so it's correct to go from this to $$ 2\cos\theta = 1 \text{ or } \sin\theta = 0. $$

Wrong:$$ab=ac\Longleftrightarrow b=c$$

Correct:$$ab=ac\Longleftrightarrow\cases{b=c\\\text{or}\\a=0}$$

Therefore, if you want to simplify by $a$, you automatically end up with a proof by exhaustion, with at least two cases.

You only consider solutions for $\cos\theta=\frac{1}{2}$ in the first case. In fact if you perform such cancellation, you should consider also solutions given by $\sin\theta=0$.

Why? Multiply both sides by zero:$$2\cos\theta=1,\,\,2\cos\theta\,\cdot0=1\,\cdot0$$ Bare in mind $\sin\theta$ could be $0$, you have $$2\cos\theta\sin\theta=\sin\theta$$

Division is valid if $\sin\theta\ne0$; however, the equation holds if $\sin\theta=0$. This is why you've to consider the case: you've assumed $\sin\theta\ne0$ at cancellation, which is correct; but the equation doesn't have this assumption. You have to consider the case when this assumption is waived.