Solve the ODE $y'^{2}+y^2=1$ I can figure out solutions $y=\sin x$ or $\cos x$ and trivial solution $y=1$. But how to get all solutions?

Failed attempt:
As the $(y,y')$ trajectory is a circle, we can write the solution as $$y(x)=\sin(u(x)),y'(x)=\cos(u(x)).$$ This requires
$$y'(x)=\cos(u(x))=u'(x)\cos(u(x))$$ and $u'(x)=1$ or $\cos(u(x))=0$.
The first case yields $u(x)=x+c$, or $y(x)=\sin(x+c)$, and the second is $y(x)=\pm1$.
Unfortunately, this method does not yield the most general solution because it does not cover the case
$$y(x)=\cos(u(x)),y'(x)=\sin(u(x))$$ that results in $u'(x)=-1$.
