Fixed points of a nonlinear system

Find the fixed points of a nonlinear two-dimensional system:

$$\dot{x} = \sin y$$ $$\dot{y} = x - x^3.$$

I know that $0 = x(1 - x²) \implies x = 0, 1, -1$. I am not sure what to do after this.

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By fixed point, I think you mean the equilibrium points of the system, right? Then simply solve $\dot{x}=0$ and $\dot{y}=0$. Any solution would be an equilibrium. In your OP, there are infinite ones, such as $y=k\pi$ and $x=0,1,-1$. Of course, the stability of different equilibriums may be different. That would be another story. –  Shiyu Dec 7 '12 at 1:20

As usual for the system of differential equations to find its fixed points you need to solve the equation $$\mathbb f(\mathbb {\tilde x}) = \mathbb 0$$ In your case it looks like
$\left\{\begin{array}{rcc} \sin y & = & 0 \\ x-x^3 & = & 0 \end{array}\right. \quad \Longrightarrow \left[ \begin{array}{ccl} y & = & \pi k,\ k \in \mathbb Z \\ x & = & \{-1,0,1\} \end{array} \right.$
Using the definition of a fixed point/equilibrium/steady state/etc. we have $$x(1 - {x^2})\mathop = \limits^{{\text{set}}} 0 \Rightarrow x = 0, \pm 1$$ and $$\sin y\mathop = \limits^{{\text{set}}} 0 \Rightarrow y = n\pi, n \in {\mathbb Z}.$$ That is all you need to do.
How common is your $\mathop = \limits^{{\text{set}}}$ notation? It looks useful. –  Jacob Dec 7 '12 at 3:29
I have seen it a few times, but it is not common. I like it because we are actually "setting" $\dot x$ and $\dot y$ to zero. Writing $\dot x = 0$ and $\dot y = 0$ is ambiguous. It is also useful in optimization when you are setting derivatives to zero. –  glebovg Dec 7 '12 at 4:15