Hence the situations are when $a$ takes the values $0,$ $1/2,$ and $1$.

According to [@ST Section II.4 equation (4.46)], the balayage measure of $\sigma=\delta_{\zeta}$ on $[-1,\,1]$ has the following differential form $$\label{zetadeltabalayage}
\mathrm{d} \widetilde{\sigma}(t)=\frac{\sqrt{\zeta^2-1}}{\pi (\zeta-t)\sqrt{1-t^2}} \, \mathrm{d}\, t.$$ We study the function $$I_a(x)=(1-a)\pi \int^1_{x} \mathrm{d} \widetilde{\sigma}(t)=(1-a) \sqrt{\zeta^2-1} \int^1_{x} \frac{\mathrm{d}\, t}{(\zeta-t)\sqrt{1-t^2}}.$$ Taking the change of variables $t=\cos \theta$ and taking into account $\zeta>2$ ($\varphi(\zeta)>2$ implies that $\arg (1-\varphi(\zeta))=\pi$), we have that $$I_a(x)=(1-a)\, \left( -\arccos x +2 \arg \left(e^{i \arccos x}-\varphi(\zeta)\right)-2\pi \right).$$ In this situation the condition of convergence (\[maincond\]) in Theorem \[main\] acquires the following form $$\label{mainconddelta}
\left|\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right] +\frac{2j-1}{2n}\pi\right|\leq A  e^{-\ell n}.$$ Then a scheme $\displaystyle {\bf x}=\left\{{\bf x}_n=\left(x_{1,n}, \ldots, x_{n,n}\right)\right\}_{n\in \Lambda}$ that satisfies the following relation is convergent $$\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]=-\frac{2j-1}{2n}\pi-A e^{-\ell n}=\kappa_{j,n},$$ with $A>0$ and $\ell>0$. This means that $$\cos \left(\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right)=\cos \kappa_{j,n}.$$ Using the cosine addition formula we have that $$x_{j,n}\cos \left\{2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}$$ $$\label{generalexample}
-\sqrt{1-x_{j,n}^2}\sin \left\{2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}=\cos \kappa_{j,n}.$$

First we consider the situation $a=1$. In this case the expressions in (\[generalexample\]) become $$\label{aonesol}
x_{j,n}=\cos \kappa_{j,n}, \quad j=1,\ldots, n, \quad 