Calculating stability of fixed points in a 2D vector flow

Question

The system in question is the following: $$ f \left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} y \\ -\mu y - k\sin{x} \end{bmatrix} $$ Given $k > 0, \mu > 0$. This of course models the motion of a simple pendulum with resistance $\mu$. The problem is to find the stability of fixed points. Evidently, there are fixed points precisely at $ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \pi n \\ 0 \end{bmatrix} $, $n \in \mathbb{Z}$. Then we may find the Jacobian: $$ J = \begin{bmatrix} 0 & 1 \\ -k \cos{x} & -\mu \end{bmatrix} $$ Which has eigenvalues, $$ \frac{\mu \pm \sqrt{ \mu^2 - 4k \cos{x} }}{2}$$ This results in the respective eigenvalues, $$ \frac{\mu \pm \sqrt{\mu^2 - 4k}}{2} \qquad \qquad \frac{\mu \pm \sqrt{\mu^2 + 4k}}{2}$$ at each fixed point. But this seems to be a problem, as this would imply that each fixed point has at least one positive real eigenvalue. However, when graphed, the field seems to have a sink at, for instance, $x = 0$. Moreover, this makes intuitive sense, as we would expect the pendulum to attract towards having angle 0. So then why does this value not result on two eigenvalues with negative real parts? What did I do wrong?