I'm trying to find the equilibrium points of $$ \begin{cases} \dot{x}=-1-y-e^{x} \\ \dot{y}=x^2+y(e^x-1) \\ \dot{z}=x+\sin(z) \\ \end{cases}$$
My try:
$$\dot{x}=0\Rightarrow -1-y-e^x=0\Rightarrow y=-1-e^x.$$
Replacing in $\dot{y}$,
$$x^2+(-1-e^x)(e^x-1)=0\Rightarrow x^2+1-e^{2x}=0\Rightarrow x=0\Rightarrow y=-2 $$ Replacing in $\dot{z}$,
$$\sin(z)=0\Rightarrow z=n\pi, n\in\mathbb{Z}. $$
So, for all $n\in\mathbb{Z}$, $(0,-2,n\pi)$ is a equilibrium point. But I don't know if exists another equilibrium points. What can I do?