One can determine equilibrium solutions of autonomous EDO's setting the derivative equal to zero. So, in this case, all the equilibrium solutions take the form $x^* = \pm \sqrt{\frac{\pi}{2} + k\pi}$, $k\in \mathbb{Z}$, right?
Well, good, now I have to determine their stability. IMHO, we can separate the analysis in two cases: when $k$ is even and when it's odd. In the first loop, if we get, say, $k=0$ we have $x^* = \sqrt{\pi/2}$. If $x<x^*$, then $\cos(x^2)>0$. In the other hand, if $x>x^*$, $\cos(x^2)<0$. So, the "$k$-even" equilibria should be all stable.
If we proceed the same way for an odd $k$, say $k=1$, we get quite the opposite and the "$k$-odd" equilibria should be all unstable.
But I have two questions regarding my "solution". 1) Can i just ignore that some arcs are negative (I've just analyzed positive ones because they all go squared so it should make no difference in my opinion)? 2) The conditions about the derivatives that I used to characterize the stability are ok, right? Thanks a lot in advance!