$T$ and $x_T-x_0$? no extremals of $J[x]= \int_0^T \frac{m}{2} \dot x^2 - \frac{k}{2}x^2 dt$ in $S=\{x \in C^1([0,T]): x(0)=x_0,x(T)=x_T\}$

Question

Functional: $$J[x]= \int_0^T \left( \frac{m}{2} \dot x^2 - \frac{k}{2}x^2 \right) dt$$ where $ \dot x = \frac {dx}{dt}$. The equation of Euler-Lagrange coincides with the Newton equation: $$m \ddot x = -kx, \\ m,k>0.$$ Determine the extremals that belong to the set: $$S=\{x \in C^1([0,T]): x(0)=x_0,x(T)=x_T\}$$ The extremals I obtained are of the form: $$x(t)=A\cos\left(\sqrt\frac{k}{m}t\right)+B\sin\left(\sqrt\frac{k}{m}t\right), A,B \in \mathbb{R}.\\$$ Using the conditions in the set: $$A=x_0. \\$$ If $T=\sqrt\frac{m}{k}n\pi, n\in \mathbb{N_0}=\{0,1,2,...\}$, then: $$x(t)=x_0\cos\left(\sqrt\frac{k}{m}t\right)+B\sin\left(\sqrt\frac{k}{m}t\right), B \in \mathbb{R}.\\$$ In this case, there are infinite extremals in $S.$ $$\\$$ Now I assume $T\neq \sqrt\frac{m}{k}n\pi, n\in \mathbb{N_0}=\{0,1,2,...\}$. The extremals are: $$x(t)=x_0\cos\left(\sqrt\frac{k}{m}t\right)+\left[x_T\csc\left(\sqrt\frac{k}{m}T\right)-x_0\cot\left(\sqrt\frac{k}{m}T\right)\right]\sin\left(\sqrt\frac{k}{m}t\right)\\$$ where $\csc(y)=\frac{1}{\sin(y)}$ and $\cot(y)=\frac{\cos(y)}{\sin(y)}.$ $$\\$$ I have a sugestion in this exercise: "Depending on $T$ and $x_T-x_0$ there are $0$ or $1$ or $\infty$ extremals." I don't understand how there is a case where there are no extremals. It must be related to $x_T-x_0$. The only thing I can think of is if the function is continuous on the interval or not, but I think it is, no matter what are the values of $x_0, x_T$.