<s>[INST] Calculate equilibrium points of a 2nd order system The following 2nd order system is given:
$$
\frac{d\textbf{x}}{dt}=\begin{bmatrix} -6 & -\frac{2}{\pi} \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix}\sin(\frac{\pi}{2}-x_2) \\ x_1x_2\end{bmatrix}
$$
The task is to calculate all equilibrium points, which I started as follows. I wrote the system down as two equations:
$$
I: -6x_1-\frac{2}{\pi}x_2+\sin(\frac{\pi}{2}-x_2)=0 \\
II: \frac{1}{2}x_2+x_1x_2=0
$$
We can see that the second equation is satisfied if either $x_2=0$ or $x_1=-\frac{1}{2}$. When I plug that into equation one I get:
$$
I, x_2=0: \\ -6x_1+\sin (\frac{\pi}{2}) = 0\\
-6x_1+1=0\\
x1=\frac{1}{6}
$$
$$
I, x_1=-\frac{1}{2}:\\ 3-\frac{2}{\pi}x_2+\sin (\frac{\pi}{2}-x_2)=0 \\
x_2=3\frac{\pi}{2}
$$
So as a result there are two equilibrium points:
$$
\textbf{x}_{R1}=\begin{pmatrix} \frac{1}{6} \\ 0 \end{pmatrix} \\
\textbf{x}_{R2}=\begin{pmatrix} -\frac{1}{2} \\ \frac{3\pi}{2} \end{pmatrix}
$$
But my problem description states that there should be four equilibrium points! How do I calculate the rest? Are the ones I already have correct?
 [/INST] What you have found is correct. But $x_2 = \pi$ and $x_2 = 2 \pi$ are also solutions in the second equations.
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