We know that if in a system there are only conservative forces, then: $$\Sigma \vec{F}=-\nabla U$$ Clearly this implies, that equilibrium points are the points in which $\nabla U=0$. Now I have some doubts that this can be applied in the case of a simple pendulum(withouth any kind of friction). In this case we have two forces acting in the system: weight and tension. Weight is clearly conservative. But is tension conservative? Well I think that in this case is conservative, since it's a vector field defined on a circle of radius equal to the lenght of the string and it's always orthogonal to this circle(so the work on any path is zero). The problem is that even if the tension vector field is conservative in this case there is no potential function for the tension! Because since the work is always zero on any path the potential should be constant, but if the potential was constant, its gradient would be zero and the tension is not $0$(I think this is due to the fact that the domain of the tension vector field is not open). And even if I was wrong and tension was not conservative, I think that we could not apply the method anyway, because it's valid only in conservative systems.
I really thought about this, but I can't find an answer. Thank you in advance.
The tension force is most directly a function of the angle and angular speed of the pendulum. It thus becomes hard to uncouple its dependence on speed or time to write it as a conservative force. Rather than considering it as a conservative force with an associated potential, we consider it to be a force of constraint given by $$\vec{Q}_r = -\bigg(mg\cos(\theta) + m\ell \dot{\theta}^2\bigg)\hat{r}\text{ .}$$
In general, an oscillator will not have conservative forces acting on it. We therefore find the equilibria of an oscillator from the equations of motion. Your oscillator is modeled as
$$ \ddot{\theta} + \frac{g}{\ell}\:\sin(\theta) =0 \text{ .}$$
Consider small oscillations $\varphi$ about an equilibrium point $\theta_e$. These oscillations given as $\theta = \theta_e + \varphi$ would obey
$$\begin{align} &(\varphi + \theta_e)'' + \frac{g}{\ell}\:\sin(\varphi + \theta_e) =0 \\ \implies &\ddot{\varphi} + \frac{g}{\ell}\:\sin(\varphi + \theta_e) =0 \\ \implies &\ddot{\varphi} + \frac{g}{\ell}\:\bigg[\sin(\varphi)\cos(\theta_e) + \cos(\varphi )\sin(\theta_e)\bigg] =0 \text{ .}\end{align} $$
In the limit as our small oscillations about $\theta_e$ tend toward zero, we must have that $\varphi$ and $\ddot{\varphi}$ are both zero. This gives us the condition
$$ \frac{g}{\ell} \: \sin{(\theta_e)} =0$$ which is solvable for the equilibrium points $\theta_e = 0,\pi$.
I'm not sure this answer is very satisfying but there is a moral to it. We shouldn't classify a force like the tension in the pendulum as a conservative force despite the fact that it "looks" conservative. We should require that a conservative force be able to do work. Forces which don't do work (or which are non-conservative) should be classified as forces of constraint. Despite the fact that we are looking at non-conservative forces, we still have tools available to us to find equilibria.