A (not so?) simple calculus problem I'm the teaching assistant for a first semester calculus course, and the professor has given the students the following problem:

Find the points on the curve $xy=\sin(x+y)$ that have a vertical tangent line.

Here is a picture of the curve:

My attempt to solve this problem led me to finding points on the given curve which also satisfy $x=\cos(x+y)$, but I can't figure out how to simultaneously solve these equations (without resulting to numerical methods, which the students are not assumed to know).  Is there something I'm missing, or has the professor given the students a problem more difficult than he intended?
Edit: I'm still looking for a solution not requiring numerical methods, or proof that no such solution exists.
 A: Let's consider $x$ as a function of $y$. We have to find points
$\left(x, y\right)$ where $x' = 0$. That yields the equation
$x = \cos\left(x + y\right)$. Then, we have a system of two equations:
$$
\left\{%
\begin{array}{rcl}
xy & = & \sin\left(x + y\right)
\\
x & = & \cos\left(x + y\right)
\end{array}\right.
$$
From those equations we get $\left\vert xy \right\vert \leq 1$ and
$\left\vert x \right\vert \leq 1$. In addition, we have
$x^{2}\left(y^{2} + 1\right) = 1$. The last identity is satisfied by the choice
$x = \cos\left(\theta\right)$ and $y = \tan\left(\theta\right)$. That means
$$
\phi
\equiv
\theta + 2n\pi = \cos\left(\theta\right) + \tan\left(\theta\right)\,,
\qquad
n \in {\mathbb Z}
$$
We have reduced the whole problem to a one variable problem:
$$
\cos^{2}\left(\phi\right)
-
\phi\cos\left(\phi\right) + \sin\left(\phi\right)
=
0
\quad\mbox{with}\quad
\left\vert%
\begin{array}{rcl}
\,\,\, x & = & \cos\left(\phi\right)
\\
\,\,\, y & = & \tan\left(\phi\right)
\end{array}\right.
$$
Solutions of this equation require
$$
\sin\left(\phi\right) \leq {\phi^{2} \over 4}
$$
Below, we can see plots of
$\cos^{2}\left(\phi\right)
 -
 \phi\cos\left(\phi\right) + \sin\left(\phi\right)$. We have to choose the roots which are consistent with the original equations since we have to pay attention to the solution signs. For example, the first positive root is
$\phi \approx 4,50321873398481\ldots$ which yields
$x \approx -0.207648303505316\ldots$ and
$y \approx 4,710867037490116\ldots$.

A: (not a complete solution, but I wanted to add a graph)
I suspect that it's a much more difficult question than intended.

By implicit differentiation, $x \frac{dy}{dx} + y = \cos( x+y) ( 1 + \frac{dy}{dx} ) $.
This gives $\frac{dy}{dx} = \frac{-y + \cos(x+y) } { x - \cos (x+y) }$.
We want a vertical tangent, which means that the denominator is 0, so $x = \cos (x+y)$.
Dividing the first equation, we get $ y = \tan (x+y)$.
There are actually infinitely many solutions, (at least) 1 for each $ n\pi < x+y < (n+1) \pi$.
The blue line is bounded between $-1 $ and $1$ and bounces back and forth. The red lines are similar to a tilted $\tan \theta$ graph.

