Very stuck with flow on a circle dynamical system problem I want to find the bifurcation points and be able to draw and classify the phase portrait as the parameter u varies for the following
$$\theta^{\bullet}=u+\sin(\theta)+\cos(2\theta)$$
(where $\theta^{\bullet}=\frac{d}{dt}\theta$)
but I am having trouble. 
This is my work and thoughts so far,
I know that fixed points occur when $\theta^{\bullet}=0$
That is when 
$u+\sin(\theta)+\cos(2\theta)=0$
I also know $\cos(2\theta)=\cos^{2}(\theta)-\sin^{2}(\theta)$
$u+\sin(\theta)+\cos^{2}(\theta)-\sin^{2}(\theta)=0$
and the most I can simplify in terms of u is
$$u=\frac{-\cos^{2}(\theta)}{\sin(\theta)(1-\sin(\theta))}$$
for $\theta$ not $0$ or $\pi/2$.
And then I am stuck, so can anyone help me with this?
Should I be trying to solve for the parameter? Or should I be trying some other method
Other thoughts: Since I cant seem to solve directly is it possible I just need to try many different parameter values like u very negative, $u=-1$, $u=0$, $u=1$ and greater? Maybe that could help but it doesn't seem systematic to me so I am not confident in that method.
Thanks
 A: Consider the Jacobian of your system defined by $$\dot{\theta} = f(\theta) = u+\sin(\theta)+\cos(2\theta)$$ which we can rewrite in terms of $\sin(\theta)$ and $\cos(\theta)$ as $$f(\theta) = u+\sin(\theta)+\cos^2(\theta)-\sin^2(\theta)$$ The Jacobian is $$J(\theta) = \cos(\theta)-2\sin(2\theta) = \cos(\theta)-4\sin(\theta)\cos(\theta)$$ The system loses structural stability when this becomes $0$ at one of its equilibria (i.e. that equilibrium becomes nonhyperbolic). Let's figure out where $(f(\theta), J(\theta)) = 0$. We note that $J(\theta) = 0$ when $\sin(\theta) = \frac{1}{4}$ or $\cos(\theta) = 0$. In the first case, $f(\theta) = u+\frac{9}{8}$, so $u = -\frac{9}{8}$ is a value of $u$ at which one or more equilibria are nonhyperbolic. In the second case, $f(\theta) = u$ or $u-2$, so $u = 0$ and $u = 2$ are two more values of $u$ where one or more equilibria are nonhyperbolic. I recommend you plot some solution curves around these values of $u$ and see precisely what is happening to the equilibria there.
