A partial differential equation with trig functions Let $\xi:\mathbb{R}^3\rightarrow \mathbb{R}$ be a function $\xi(x,y,z)$. I have the following PDE:
$$
\cos(\xi)\partial_y\xi = \sin(\xi)\partial_x\xi
$$
which is equivalent to $\tan(\xi)=\xi_y/\xi_x$.
Clearly, $\xi_x = \xi_y = 0$ (where $\partial_i\xi=\xi_i$) is a solution, which would imply $\xi(x,y,z) = \eta(z)$ is only a function of $z$. 
But are there any other solutions? If not, how to prove it?
My thoughts so far: obviously, 
\begin{align}
\xi_x &= \cos(\xi)\\
\xi_y &= \sin(\xi)
\end{align} 
together describe a solution.
We can solve these separately to get:
$$
\xi = 2\arctan(\exp(y+f_1(x,z))\;\;\;\&\;\;\; 
\xi=2\arctan(\tanh([1/2][x+f_2(y,z)]))
$$
respectively. 
This implies that $$
\exp(y+f_1(x,z)) = \tanh([x+f_2(y,z)]/2) $$
Not sure if this line of thinking is useful though.
 A: This is an expanded version of what I mentioned in the comments.
In case you aren't familiar with the method of characteristics, there is a simple trick for solving this kind of equation: interchange $x$ with $\xi$.  Let $X(u,y,z)$ be the solution of $$u=\xi(X,y,z).$$  To evaluate your PDE $$\xi_y\cos\xi=\xi_x\sin\xi$$ in new variables, differentiate the above equation.  Taking $\partial_u$, it follows that $ 1= \xi_X X_u$, so $\xi_X=1/X_u$, assuming $X_u\neq 0$ of course.  Taking $\partial_y$, we have $0=\xi_XX_y+\xi_y$, so $\xi_y=-X_y/X_u$.  Substituting into your PDE, we get:
$$
-X_y(u,y,z)\cos u =\sin u,
$$
such that $X_y=-\tan u$ for $u\notin \pi (\mathbb Z+1/2)$.  We can write the general solution as $$X(u,y,z)=(f(u,z)-y)\tan u,$$
such that $\xi(x,y,z)$ is found by solving the equation
$$
x=(f(\xi(x,y,z),z)-y)\tan\xi(x,y,z).
$$
Let us rewrite this implicit solution as follows:
$$
\tan\xi(x,y,z)=\frac{x}{f(\xi(x,y,z),z)-y}.
$$
Since $f$ is an arbitrary function, let us choose $f(\xi,z)=g(z)$, for example.  Then we obtain an explicit solution:
$$
\xi(x,y,z)=\arctan\left(\frac{x}{g(z)-y}\right),
$$
where $g$ is an arbitrary function, and $(x,y,z)$ are restricted to domains where the right hand side makes sense.
A: Characteristic curves of $0=\sin(ξ)ξ_x-\cos(ξ)ξ_y$ follow the direction field
$$
\frac{dx}{\sin(ξ)}=\frac{dy}{-\cos(ξ)}=\frac{dz}{0}=\frac{dξ}{0}
$$
which means that $\cos(ξ)x+\sin(ξ)y=c_1$, $z=c_2$, $ξ=c_3$ are constant along these curves and the curves are completely characterized by these first integrals. As there are 3 constants for a family of curves in the 3 dimensional graph of $ξ$, there is one parameter too much, that is, there exists a relation $g(c_1,c_2,c_3)=0$ resp. $g(\cos(ξ)x+\sin(ξ)y,z,ξ)=0$ among them.
For example, if $u$ is any differentiable function of $2$ variables, the solution of the implicit equation
$$
ξ=u(\cos(ξ)x+\sin(ξ)y,z)
$$
satisfies the given equation as long as $$(\cos(ξ)y-\sin(ξ)x)\,\partial_1u\ne 1.$$
A: $$
\cos(ξ)\frac{\partial ξ}{\partial y}- \sin(ξ)\frac{\partial ξ}{\partial x}=0 \tag 1
$$
With the method of characteristics, the set of ODEs for the characteristic curves is :
$$\frac{dy}{\cos(ξ)}=\frac{dx}{-\sin(ξ)}=\frac{dz}{0}=\frac{dξ}{0} \quad\implies\quad dξ=0\quad\text{and}\quad dz=0$$
A first family of characteristic curves comes from $dξ=0 \quad\to\quad ξ=c_1$.
A second family of characteristic curves comes from $\frac{dy}{\cos(c_1)}=\frac{dx}{-\sin(c_1)} \quad\to\quad x\cos(c_1)+y\sin(c_1)=c_2$
A third family of characteristic curves comes from $dz=0 \quad\to\quad z=c_3$.
This is true any $c_1$ and $c_2$ and $c_3$, hence the equation $\Phi(c_1,c_2,c_3)=0$ is true any function $\Phi$ of three variables.
$$\Phi\left(ξ\:,\:x\cos(ξ)+y\sin(ξ)\:,\:z\right)=0$$
Or, on another equivalent form :
$$ξ=F\left(x\cos(ξ)+y\sin(ξ)\:,\:z\right)\tag 2$$
where $F$ is any differentiable function of two variables.
This is the general solution of the PDE $(1)$ expressed on implicit form.
If one want the explicit form $ξ(x,y,z)$, one have to solve $(2)$ for $ξ$. This is not always analytically possible because there is not always a closed form for the roots of the equation $(2)$. This depends on the function $F$.
The function $F$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, we cannot go further. 
Nevertheless, one can answer to the raised question : They are an infinity of solutions for $(1)$ : Simply put in $(2)$ any differentiable function and you get a solution. Even more, if you chose some convenient function $F$ , the equation $(2)$ will be solvable on closed form and you can get some explicit solutions $ξ(x,y,z)$.
EXAMPLE :
If we chose $F(X,z)=\arcsin(X)\qquad$ 
We are allowed to chose a function in which the second variable doesn't appear. 
$ξ=\arcsin\left(x\cos(ξ)+y\sin(ξ)\right) \quad\to\quad \sin(ξ)=x\cos(ξ)+y\sin(ξ)$
$\tan(ξ)=\frac{x}{1-y} \quad\to\quad ξ(x,y,z)=\arctan\left(\frac{x}{1-y}\right)\quad$ is a solution (one among many others).
