Can't figure this out: $dy/dx = \cos(y)\sin(x)$ Problem: Find the solution and domain of validity for the following differential equation:
$\dfrac{dy}{dx} = \cos(y) \cdot \sin(x)$
What I tried: I noticed that this is a separable equation, so I wrote it in the form: $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$
Next, I found out that if $\cos(y) = 0$ then $\dfrac{1}{\cos(y)}$ is undefined. So I separate into two following cases;
Case1: $\cos(y) \neq 0$;
Then $\int\dfrac{dy}{\cos(y)} = \int\sin(x)dx$. By solving these integrals I get that $\tan(y) \cdot \sec(y)+ c_0 = -\cos(x)$ which implies that
$\arccos(-\tan(y) \cdot \sec(y)+ c) = x$,   for $c=-c_0$ which is an arbitrary constant.
I wonder if this is a general solution.
Case2: $\cos(y)=0$;
This implies that $y\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$
Pick an arbitrary $y_0\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$, substituting this into the differential equation gives that;
$\dfrac {dy_0}{dx} = \cos(y_0)\cdot \sin(x)$.
I am pretty sure that $\cos(y_0) = 0$. But I do not know how can I show that $\dfrac {dy_0}{dx} = 0$ to find out if there is a singular/lost solution or not.
Moreover, I have been taught that $dx$ means derivative, for instance $x^2\cdot dx = 2x$. Then why can't we solve a separable differential equation just by taking derivative of both sides for instance in $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$

$$\int\frac{dy}{\cos(y)}=\int\frac{\cos(y)\,dy}{1-\sin^2(y)}=\text{artanh}(\sin(y))=\int\sin (x)\,dx$$
gives
$$y=\arcsin(\tanh(c-\cos(x)))$$ and periodic replicas.
$$y=\pm\dfrac\pi2$$ and replicas are also (degenerate) solutions ($y'=0$, corresponding to $c=\pm\infty$).

