How do I solve $y' = \sin(x - y)$? How can I solve the differential equation:
$$y'=\sin(x-y)$$
Could I do this?
$$\frac{dy}{dx}= \sin x \cos y - \sin y \cos x$$
But, how would I continue?

I will only add the solution of the integral,
$$\int\dfrac{1}{1-\sin x}dx$$ set $\tan(x/2)=u$ then you will have
$$\int\dfrac{1}{1-{2u\over{1+u^2}}} \dfrac{2}{u^2+1} du$$ 
As a result you will have $$\dfrac{2\sin(\dfrac x2)}{\cos(\dfrac x2)-\sin(\dfrac x2)}+c$$
