I'm really struggling to find the bifurcation(s) of the system
$x'=x^2 + \cos(x+ \mu)$, $\mu \in [0,2\pi)$.
I've tried substituting $y=\mu+x$, taylor expanding, and just about everything else I could think of!
I know that the requirement for bifurcation is $f(x,\mu)=f'(x,\mu)=0$ but I just can't work out any solutions. Maple says the actual answer is $0.486,1.323$ but I can't seem to work out how to get there.
Thanks very much in advance and I hope you have better luck than I did!
I found one of the answers you gave by finding two equations in $\mu$:
First I rewrote $f(x) = 0$ as $\mu = \arccos(-x^2) -x$ and $f'(x) = 0$ as $\mu = arcsin(2x) - x$.
Then I equated the two:
$$\arccos(-x^2) - x = \arcsin(2x) -x$$ $$\arccos(-x^2) = \arcsin(2x)$$ $$\frac{\sin(\arccos(-x^2))}{2} = x$$
From which I derived the iterative formula: $$x_{n+1}= \frac{\sin(\arccos(-x_n^2))}{2}$$
Which gives the answer $x = 0.48587$ to 5 decimal places when $x_0 = 0$.
P.s. I realise that I may have ignored the other answers since $\sin(x) = \sin(\pi - x)$ and $\cos(x) = \cos(-x)$ but from my working they all gave the same answers.