Fixed points of: $\dot{x}=\sin(y) \qquad \dot{y}=\cos(x)$ How can you find the fixed points of this system:
$\dot{x}=\sin(y)\\ 
\dot{y}=\cos(x)$
Normally I would suggest that you find the points when both functions are equal to 0.
 A: As you suggested, to solve the fixed points simply set $\dot{x} = \dot{y}= 0$.
So $$\dot{x} = 0 \implies y = n\pi\quad \forall n \in \mathbb{N},$$
and $$\dot{y} = 0 \implies  x = k\pi + \frac{\pi}{2} \quad \forall k \in \mathbb{N}.$$
Hence your fixed points are all the pairs $(x,y) = \left(n\pi+\frac{\pi}{2}\pi,k\pi\right)$ for $n,k\in\mathbb{N}.$
A: Maybe this helps
\begin{eqnarray*}
\dot{x} &=&\sin y,\;\dot{y}=\cos x \\
\dot{x}\cos x &=&\dot{y}\sin y=\cos x\sin y
\end{eqnarray*}
so
\begin{eqnarray*}
\dot{x}\cos x-\dot{y}\sin y &=&0 \\
\partial _{t}\sin x &=&\dot{x}\cos x,\;\partial _{t}\cos y=-\dot{y}\sin y \\
\partial _{t}\{\sin x+\cos y\} &=&0 \\
\sin x+\cos y &=&C
\end{eqnarray*}
A: No, you're thinking of stationary points. I'm guessing this question is to do with the Banach contraction mapping principle. You would rewrite the coupled DEs in the form:
$$X(t)\stackrel{def}{=}\left(\begin{array}{c}x(t)\\y(t)\end{array}\right) = \int_0^t \left(\begin{array}{c}\sin(y(u))\\\cos(x(u))\end{array}\right)\mathrm{d}u$$
i.e. you write it in the form $X = \mathscr{L} X$, where $\mathscr{L}:C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]\to C^1[-t_{max},t_{max}]\times C^1[-t_{max},t_{max}]$ is the integral operator defined above. Then you prove that $\mathscr{L}$ is contractive: i.e. $\exists N\geq0,\,0<K<1$ such that $\|\mathscr{L}^N X_1-\mathscr{L}^N X_2\| \leq K \|X_1-X_2\|$ for all $X_1,\,X_2\in C^1[-t_{max},t_{max}]$. Under these conditions, repeated iteration of $\mathscr{L}$ on any "point" (i.e vector of functions) $X_0\in C^1[-t_{max},t_{max}]$ will converge to the unique fixed point $X_\infty\in C^1[-t_{max},t_{max}]$ such that $\mathscr{L} X_\infty = X_\infty$. In effect, I think you're being asked to work through the proof of the local version of the Picard-Lindelöf Theorem for this differential equation. The inequalities $|\sin(x_1-x_2)| < |x_1-x_2|$ and $|\cos(x_1-x_2)| < |x_1-x_2|$ will help you. Also, use the maximum modulus ($\mathcal{L}^\infty$) norm to define the metric space of functions.
You're probably not being asked to explicitly give a closed form for the fixed point (i.e. the solution of the DE), but simply to prove it exists and is unique by the arguiments shown to you on the Wiki page.
