Question: How can I solve $y-xy'-\sin(y')=0$? Are there any general techniques for solving ODE of the form $y=f(y')$, where $f$ is a trigonometric function?

Answer: 
Assuming you mean $y(x)$, what you have is what is known as a [Clairaut's equation](http://en.wikipedia.org/wiki/Clairaut%27s_equation).

Separate your $y$ and $y'$ and then differentiate with respect to $x$, then factor and solve.

After a bit of manipulation, you should get $y = \sin \alpha + \alpha x$ or $y = \pm(x \arccos (-x) + \sqrt{-x^2 +1})$.
I really don't know how to solve this kind of ODE's in a general manner. For your specific case, basically by trial and error, I found y = a x + Sin[a]. Please, let me know if you receive answers. I am really curious.