Find all the critical points and determine the phase plane solution curves for the system

\(\begin{array}{c}\frac{{dx}}{{dt}} = sinxcosy,\\\frac{{dy}}{{dt}} = cosxsiny\end{array}\)

To find the critical points we have to solve the system\({\bf{x' = 0,y' = 0}}\), so one has;

\(\begin{array}{c}{\bf{0 = sinxcosy}}\\{\bf{0 = cosxsiny}}\end{array}\)

The first equation is satisfied for \({\bf{x = }}k\pi \) or \({\bf{y = }}\frac{{\bf{\pi }}}{{\bf{2}}}{\bf{ + k\pi ,k}} \in {\bf{Z}}\) and the second equation is satisfied for \(x{\bf{ = }}\frac{\pi }{2}{\bf{ + }}l\pi \) or\(y{\bf{ = }}l\pi ,l \in \mathbb{Z}\), so they are both satisfied for

\({\bf{x = k\pi ,y = l\pi }}\) Or \({\bf{x = }}\frac{{\bf{\pi }}}{{\bf{2}}}{\bf{ + k\pi ,y = }}\frac{{\bf{\pi }}}{{\bf{2}}}{\bf{ + l\pi ,}}\,{\bf{k}} \in Z\)

The critical points are given by \((x,y)\mid x = k\pi ,y{\bf{ = }}l\pi \;or\;x{\bf{ = }}\frac{\pi }{2} + k\pi ,y{\bf{ = }}\frac{\pi }{2}{\bf{ + }}l\pi ,k,l \in \mathbb{Z}\)

Now, one can solve for the phase plane solution curves.

The related phase plane equation is:

\(\frac{{dy}}{{dx}}{\bf{ = }}\frac{{\cos x\sin y}}{{\sin x\cos y}} \Rightarrow \cot ydy{\bf{ = }}\cot xdx\)

Integrating the previous equation, one will obtain the phase plane solution curves:

\(\begin{array}{c}\int {\cot } ydy{\bf{ = }}\int {\cot } xdx\\\ln \left| {\sin y} \right| = \ln \le \left| {\sin x} \right| + C\\\sin \left| y \right|{\bf{ = }}|\sin x|{e^C}\\\sin \left| y \right| = \lambda |\sin x|\end{array}\)

Therefore, phase plane solution curves are given by\(|\sin y|{\bf{ = }}\lambda |\sin x|\)