for which values of $\theta$ does this equation $x_{n+1}=\cos(\theta)x^2_{n}-\sin(\theta)x^2_{n-1}$ have bounded solutions? I would like to investigate the global behavior of the following equation :
$$x_{n+1}=Ax^2_{n}-Bx^2_{n-1}$$
where $A(\theta)= \cos(\theta)$ and $B(\theta) =\sin(\theta)$  are nonnegative parameters and $x_{-1}$, $x_0$ are nonnegative with $x_{-1}+x_{0}>0$. Here is my question:


*

*For which values of $\theta$ does the equation below have bounded solutions?
$$x_{n+1}=\cos(\theta)x^2_{n}-\sin(\theta)x^2_{n-1}$$ 


Note : $\theta$ $\in \mathbb{R}$, $n=0,1,2\cdots$.
Thank you for any help.
 A: I ran a small computation: The following plot is created as follows:
For each point $r(\cos \theta, \sin \theta)$, I use $x_{-1} = 0$, $x_0 = r$
as initial values, and $\theta$ as the parameter.
The white area is where the iterations is less than $2$ (bounded), the first 20 iterations. As we see, a small circle around the origin is white, meaning that there are small initial values that is bounded for every $\theta$. Now, this is not a proof, but the picture suggests this.
Changing the cut-off to 40 iterations does not change the picture much.

EDIT: I think I have a proof of the statement. 
Let $r=1/\sqrt{2}$. If $x_{-1},x_0 \leq r$, then the sequence is bounded for all $\theta$.
Proof: $|\cos(\theta)x_{n-1}^2 - \sin(\theta)x_{n}^2| \leq r^2(|\cos \theta|+|\sin \theta|) \leq r^2 \sqrt{2} \leq 1/\sqrt{2}$, and the statement follows from induction.
ValueFrom[x_, y_] := Module[{func, xf, xfp, iterTot, t, r, cc, sc},
   t = Arg[x + I y];
   r = Abs[x + I y];

   cc = Cos[t];
   sc = Sin[t];

   func[{xn_, xp_, i_}] := {xp, cc*xp^2 - sc*xn^2, i + 1}; 

   {xf, xfp, iterTot} = 
    NestWhile[func, {0, r, 0}, Abs[#[[2]]] < 2 && #[[3]] < 20 &];
   iterTot
   ];

DensityPlot[ValueFrom[x, y], {x, -2.0, 2.0}, {y, -2.0, 2.0},
 PlotPoints -> {150, 150}, PlotRange -> All, 
 ColorFunction -> "SunsetColors"]

