TITLE: Differential equation $\cos(f')=\cos(f)'$
QUESTION [5 upvotes]: The differential equation
$$\sin(f')=\sin(f)'$$
has the trivial solution $f=0$. Does the equation
$$\cos(f')=\cos(f)'$$
have any solutions?

REPLY [6 votes]: These are two separable equations.
From
$$\sin(f')=\cos(f)f'$$
we deduce
$$\text{sinc}(f')=\cos(f)$$and the non-trivial solution
$$\int\frac{df}{\text{sinc}^{-1}(\cos(f))}=x+C.$$
And from $$\cos(f')=-\sin(f)f',$$
defining $g(t)=\cos(t)/t$ (there is no standard cardinal cosine), we have
$$g(f')=-\sin(f),$$and
$$\int \frac{df}{g^{-1}(-\sin(f))}=x+C.$$
As the functions are not univocally invertible, several branches are possible.