Q:

Stability (solution?) of nonlinear difference equation

I'm trying to solve the following difference equation:
$$x_n = x_{n - 1} + 2\pi \lambda \sin(x_{n-1})$$ 
I'm an electrical engineer, so I only briefly learned about difference equations. I'm looking for a solution, but more important, I need to show that the equation is instable for some values of $\lambda$. 
In my analysis, I've found the value of $\lambda$, for which the equation becomes unstable, $\lambda = \pi^{-1}$. I showed this by assuming that $x_n = 2\pi - x_{n - 1}$, which can also be seen in a simple matlab simulation. I'm not entirely happy with this reasoning, because I'm not sure if I'm missing out on solutions that do not fit this premise.
How do you do this the proper way?

A:

Let $f(x) = x + 2 \pi \lambda \sin(x)$.  There are fixed points $n \pi$ where $f(x)=x$ corresponding to constant solutions.  A fixed point $p$ is unstable if $|f'(p)| > 1$ and
stable if $|f'(p)| < 1$. 
In this case $f'(n\pi) =  1 + 2 \pi \lambda (-1)^n$, so for $n$ odd the fixed point $n \pi$ is stable for $0 < \lambda < 1/\pi$ and
for $n$ even it is stable for $-1/\pi < \lambda < 0$.  Of course for $\lambda = 0$ the recurrence is just $x_{n+1}=x_n$ and all points are stable fixed points.
But "stable" and "unstable" applies to solutions, not the equation.  When a solution becomes unstable, another one may take its place.  In this case when $\lambda$ increases past $1/\pi$, it appears that you get stable $2$-cycles for a while, then a period-doubling cascade and then mostly chaos, which is typical for this kind of dynamical system.

