Trigonometric differential equation Is it possible to solve the following ordinary differential equation:
$\theta'(t)=x(t)\sin(\theta(t))-y(t)\cos(\theta(t)),\ \forall t\in I$, $I-$ interval from $\mathbb{R}$,
where $x,y:I\to\mathbb{R}$ are two given continuous functions?
 A: A simplificationcan be obtained in the following way: Write
$$\bigl(x(t),y(t)\bigr)= r(t)\bigl(\cos\alpha(t),\sin\alpha(t)\bigr)\ ,$$
where the functions $t\mapsto r(t)>0$ and $t\mapsto\alpha(t)$ can be considered as given. The ODE then becomes
$$\theta'(t)=r(t)\>\sin\bigl(\theta(t)-\alpha(t)\bigr)\ .\tag{1}$$
Now introduce a new independent variable $\tau$ by means of
$$\tau:=\psi(t):=\int_0^t r(s)\>ds\ ,$$
and denote differentiation with respect to $\tau$ by a dot. The ODE $(1)$ then becomes
$$\dot\theta(\tau)=\sin\bigl(\theta(\tau)-\tilde\alpha(\tau)\bigr)\ ,$$
where $\tilde\alpha(\tau):=\alpha\bigl(\psi^{-1}(\tau)\bigr)$ can again be considered as given.
A: This is not a full answer to the question but it seems to be too long for a comment.
The ordinary differential equation you posted seems incredibly complex and I really wonder if there is any way to approach a solution.
I tried using $x(t)=a$ and $y(t)=b$ just to see the kind of result we could get. Even for a CAS, the problem does not look very simple and the answer is $$\theta(t)=2 \tan ^{-1}\left(\frac{\sqrt{a^2+b^2} \tanh \left(\frac{1}{2} \sqrt{a^2+b^2}
   \left(c-t\right)\right)-a}{b}\right)$$
I later tried  using $x(t)=a+bt$ and $y(t)=c+dt$ and I (the CAS !) did not arrive anywhere. Any other, even simple, functions I tried led to the same lack of success.
