How to linearize this nonlinear system?

Question

Take a look at this nonlinear system $$ \dddot{x} +4\ddot{x}+24|\dot{x}| + 5\cos(x)|\dot{x}| + 50x = u $$ The objective is to linearize the system about the equilibrium point. First, we compute the equilibrium point but we need first to convert the third degree of ODE to first order ODE. $$ \begin{align} x_1 &= x \implies \dot{x}_1 = x_2 \implies f_1(\boldsymbol{x},u) \\ x_2 &= \dot{x} \implies \dot{x}_2 = x_3 \implies f_2(\boldsymbol{x},u) \\ x_3 &= \ddot{x} \implies \dot{x}_3 = \dddot{x} \implies f_3(\boldsymbol{x},u) \\ \end{align} $$ where $$ \dddot{x} = u - 4x_3-24|x_2| - 5\cos(x_1)|x_2| - 50x_1 $$ The new form of the system is $$ \begin{align} \begin{bmatrix} f_1(\boldsymbol{x},u) \\ f_2(\boldsymbol{x},u) \\ f_3(\boldsymbol{x},u) \end{bmatrix} &= \begin{bmatrix} x_2 \\ x_3 \\ u - 4x_3-24|x_2| - 5\cos(x_1)|x_2| - 50x_1 \end{bmatrix} \\ \dot{\boldsymbol{x}}&=f(\boldsymbol{x},u) \end{align} $$ The new form is indeed a first order system. Assume $u=0$. Now, we compute the equilibrium point(s) which can be obtained as follows: $$ \begin{align} \dot{\boldsymbol{x}}&=f(\boldsymbol{x}_e,u_e) = 0 \tag{1} \end{align} $$ The only equilibrium point for Eq.(1) is $(\boldsymbol{x}_e,u_e)=(\boldsymbol{0},0)$ or $x_1=0,x_2=0,x_3=0,u=0$. Now we compute the Jacobain matrix as follows: $$ \begin{align} A = \frac{\partial f(\boldsymbol{x},u)}{\boldsymbol{x}} \Big|_{\boldsymbol{x}=\boldsymbol{x}_e} &= \begin{bmatrix} \frac{\partial f_1(\boldsymbol{x},u)}{x_1} & \frac{\partial f_1(\boldsymbol{x},u)}{x_2} & \frac{\partial f_1(\boldsymbol{x},u)}{x_3} \\ \frac{\partial f_2(\boldsymbol{x},u)}{x_1} & \frac{\partial f_2(\boldsymbol{x},u)}{x_2} & \frac{\partial f_2(\boldsymbol{x},u)}{x_3} \\ \frac{\partial f_3(\boldsymbol{x},u)}{x_1} & \frac{\partial f_3(\boldsymbol{x},u)}{x_2} & \frac{\partial f_3(\boldsymbol{x},u)}{x_3} \end{bmatrix} \\ &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -50 & \frac{24x_2+5\cos(x_1)x_2}{|x_2|} & -4 \end{bmatrix} \end{align} $$ Clearly $\frac{24x_2+5\cos(x_1)x_2}{|x_2|}$ is undefined at the equilibrium point. How to proceed from this point?