Answer
See below
Work Step by Step
We obtain:
$4x-y-y\sin x=0\\
x+2y=0$
The critical point is $(0,0)$
The Jacobian of the system is:
$J(x,y)=\begin{pmatrix}
4-y\cos x & -\sin x-1\\
1 & 2
\end{pmatrix}$
Substituting:
$J(0,0)=\begin{pmatrix}
4 & -1\\
2 & 1
\end{pmatrix}$
Then the eigenvalues are $\lambda_1=\lambda_2=3$.
Consequently, the equilibrium point $(0,0)$ is a node.
