= 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 to
$$
(\cot x+y)dx+xdy=d[\ln|\sin x|+xy].
$$
Therefore the implicit solution family is $xy+\ln|\sin x|=C$. On each graph <interval> between consecutive multiples of $\pi$, this becomes
$$
\boxed{y(x)=\frac{C-\ln|\sin x|}{x},\qquad k\pi<x<(k+1)\pi,\quad k\in\mathbb Z.}
$$
No 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> \b[$x=k\pi$] are additional solutions: along them $dx=0$ and $x\sin x\,dy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an <interval>.
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