Answer
See below
Work Step by Step
We obtain:
$2y+\sin x=0\\
x(\cos y-2)=0$
The critical point are $(0,0)$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
\cos x & 2\\
\cos y-2 & -x\sin y
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
1 & 2\\
-1 & 0
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=\frac{1}{2}+\frac{i\sqrt 7}{2},\lambda_2=\frac{1}{2}-\frac{i\sqrt 7}{2}$.
Consequently, the equilibrium point $(0,0)$ is a spiral.
