Equilibrium Points Second Order Differential 
Attempt:
I get the system of the two first order equations (first order in $w$) by considering the different signs the first derivative takes. Problem is by equilibrium points: do I just set the first derivative to 0 in the (*) equation or do I get them from the two first-order equations? I just need someone to tell me how to obtain the equilibrium points. Thanks. 
 A: One way to find the equilibria:
set
$\omega = \dfrac{d\theta}{dt} = \dot \theta; \tag{1}$
then
$\dot \omega = \ddot \theta = -c\omega \vert \omega \vert - \sin \theta, \tag{2}$
since
$\ddot \theta + c\dot \theta \vert \dot \theta \vert + \sin \theta = 0. \tag{3}$
At an equilibrium of the system (1)-(2) and/or (3), we have $\theta(t) = \text{constant}$ for all $t$; thus by (1),
$\omega = \dot \theta = 0 \tag{4}$
at such points, so we must also have
$0 = \dot \omega = -c \omega \vert \omega \vert - \sin \theta = -\sin \theta; \tag{5}$
but
$\sin \theta = 0 \Leftrightarrow \theta = n\pi, n \in \Bbb Z; \tag{6}$
the equilibria thus occur at those points in the $\theta$-$\omega$ plane at which $\theta = n\pi$, $\omega = 0$, or equivalently, for the initial conditions $\theta = n \pi$, $\dot \theta = 0$ for equation (3).  Thus we see how the equilibria are found, and exactly what points they are; thus our OP WhizKid's primary question is answered.
Of course, the other questions posed in the problem text are very engaging in their own right, and merit a careful discission, this especially due to the $c \dot \theta \vert \dot \theta \vert$ term.  This I may take up at a later date; meanwhile, I hope to soon post on a related question, 
How to characterize the non-trivial solutions of this non-linear differential equation? Are they all periodic?, the analysis of which bears on the present problem.
