solve $y(x)=\cos \left(y'(x)\right) + y'(x)\sin (y'(x)), y(0)=1$ solve $$y(x)=\cos (y'(x)) + y'(x)\sin (y'(x)), y(0)=1$$
with wolfram alpha I got that a solution is $y(x)=x\arcsin x+\cos (\arcsin x)$
but I have no idea how to find it.
I tried transforming into an exact equation by letting $u=y'$ then I get $y=\frac{u^2}{2}$
I just realised this is wrong (because $\frac{d}{dx}\frac{u^2}{2}=u'u\neq y'$) so there's no point posting the rest of my work.
any hint?

Hint:  Apply the chain rule of calculus.
