Losing solutions when dividing by an expression There's something that's been  aggravating me when solving implicit differentiation. For instance, take the implicitly defined curve
$$\cos x+\cos y=\frac{1}{2}$$
(Differentiate both sides w.r.t x)
$$-\sin x-\sin y \frac{dy}{dx}=0$$
$$-\frac{dy}{dx}\sin y= \sin x$$
$$\frac{dy}{dx}=-\frac{\sin x}{\sin y}$$
The last step is the one which really confuses me, which when we divide by $(-\sin y)$: aren't we saying by doing this step that $(-\sin y)$ won't be equal to zero; aren't we kind of erasing or canceling certain values on the graph / function?
Like when we do in questions such as:

$3\cos x-12\sin x\cos^2x=0$,
$0<x<\pi$

We can make the equation look neater by dividing by $\cos x$ but we will lose solutions, as $\cos x$ won't be equal to zero and $x=\pi / 2$ won't be among our answers.
 A: I always had the same weird feeling when I solve elementary differential equations or differentiate implicit functions. But the following will resolve your concerns thoroughly.
If you have plotted the graph of $\cos x + \cos y = 1/2$, you could see that the derivative at $\sin y = 0$ is actually undefined (vertical). Therefore you can only differentiate assuming that $y$ is differentiable. This differentiability condition gives us the fact that $\sin y\neq 0$.
A useful theorem is the implicit function theorem. We only need the 2-dimensional result in our case. The theorem (2-dimensional) states that if $f$ is a continuously differentiable function (in your case $f(x, y)=\cos x +\cos y - 1/2$) that defines a curve $f(x, y)=0$, then at any point $(a, b)$ on the curve, there exists an open set of $U\ni (a, b)$ and a unique continuously differentiable function $g:U\to \mathbb{R}^2$ such that $y=g(x)$ if the Jacobian of the implicit function, namely
$$\frac{\partial f}{\partial y}(a, b)\neq 0$$
in our 2-dimensional case. The differentiability of $f$ ensures the existence of this partial derivative.
Now, for our $f(x, y)=\cos x +\cos y - 1/2$, we have $y$ is a continuously differentiable function of $x$ if $\frac{\partial f}{\partial y}(a, b)=-\sin y\neq 0$. This leads to the assumption $\sin y\neq 0$ directly.

The problem you had is not with factoring or moving factors. At the moment you apply the operator $\frac{d}{dx}$, you automatically assume $y$ is a differentiable function of $x$ (which requires $\sin y\neq 0$). Thus, you can write $\frac{dy}{dx}=-\frac{\sin x}{\sin y}$ at the points where $y$ is differentiable.
A: You should note that "dividing won't always work, but factorising always works"
$$(\sin(y))\cdot \frac{dy}{dx} = -\sin x$$ is always true, but
$$\frac{dy}{dx}=-\frac{\sin x}{\sin y}$$ is only true where $\sin y \neq 0$.
Likewise, if you factorise your trig equation as $\cos x (3-12\sin x \cos x) = 0$, you won't miss any solutions, because it is possible that $\cos x = 0$.
Sidenote: I encourage you to plot your graph on Desmos (or other software). It's actually kind of pretty.
