 Solution

Away from zeros of $\sin x$, an integrating factor for a differential one-form is $\mu (x)=1/\sin x$. Multiplication converts the form toTherefore the implicit solution family is $xy+\ln |\sin x|=C$. On each graph interval between consecutive multiples of $\pi$, this becomesNo such open interval contains zero. At $x=k\pi$, the undivided graph equation would read $\cos (k\pi )=0$, which is impossible for a finite derivative; hence no differentiable graph extends across one of these points.

If solutions are instead understood as arbitrary parametrized integral curves of the differential form, the vertical lines $x=k\pi$ are additional solutions: along them $dx=0$ and $x\sin xdy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an interval.