= Solution
Introduce $v=\dot x$. The <phase plane> system for the <damped pendulum> is
$$
\dot x=v,\qquad \dot v=-2kv-\omega^2\sin x.
$$
At an <equilibrium point>, $v=0$ and $\sin x=0$, so all <equilibrium points> are $\boxed{(x,v)=(n\pi,0),\ n\in\mathbb Z}$. Angles identified modulo $2\pi$ give one downward and one upward <equilibrium point>.
The <linearization of a dynamical system> at $(n\pi,0)$ has <Jacobian matrix> and <eigenvalues>
$$
J_n=\begin{pmatrix}0&1\\-\omega^2(-1)^n&-2k\end{pmatrix},\qquad \lambda_\pm=-k\pm\sqrt{k^2-\omega^2(-1)^n}.
$$
When $n$ is even and $k>\omega$, both <eigenvalues> are real and strictly negative, since $0<\sqrt{k^2-\omega^2}<k$. These downward <equilibrium points> are \b[<stable nodes>]. When $n$ is even and $k<\omega$, the <eigenvalues> are $-k\pm i\sqrt{\omega^2-k^2}$, so the downward <equilibrium points> are \b[<stable foci>], approached with decaying oscillations.
When $n$ is odd, $\sqrt{k^2+\omega^2}>k$, so one <eigenvalue> is positive and one is negative in either parameter range. Every upward <equilibrium point> is consequently an \b[unstable <saddle equilibrium>]. All these <equilibrium points> are <hyperbolic equilibria> in the two requested cases, so their <linearization of a dynamical system> gives their local nonlinear classification.
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