n\in \Lambda.$$ The nodes are close to the zeros of the Chebyshev polynomials. That’s why the term corresponding to the $\sigma$’s influence in (\[generalexample\]) vanishes when $a=1$.

Let us analyze now the case $a=1/2$. We consider the following identities $$\label{condcos}
\cos \left[\pi-\arg \left(e^{i \arccos x_{j,n}}-\varphi(\zeta)\right)\right]=\frac{\varphi(\zeta)-x_{j,n}}{\sqrt{\varphi^2(\zeta)-2 \varphi(\zeta) x_{j,n}+1}}$$ and $$\label{condsin}
\sin \left[\pi-\arg \left(e^{i \arccos x_{j,n}}-\varphi(\zeta)\right)\right]=\frac{\sqrt{1-x_{j,n}^2}}{\sqrt{\varphi^2(\zeta)-2 \varphi(\zeta) x_{j,n}+1}}.$$ Substituting (\[condcos\]) and (\[condsin\]) in (\[generalexample\]) we arrive at the quadratic equations: $$x_{j,n}^2-\frac{2\sin^2\kappa_{j,n}}{\varphi(\zeta)}\, x_{j,n}+ \frac{\sin^2\kappa_{j,n}}{\varphi^{2}(\zeta)} -\cos^2 \kappa_{j,n}=0, \quad j=1,\ldots, n, \quad n\in \Lambda.$$ For each $j=1,\ldots, n,$ $ n\in \Lambda$ we obtained the following solutions $$\label{ahalfsol}
x_{j,n}=\frac{1}{\varphi(\zeta)}\left[\sin^2 \kappa_{j,n}+ \cos \kappa_{j,n} \sqrt{\varphi^2(\zeta)-\sin^2 \kappa_{j,n}}\right].$$ During the process of finding these above solutions we introduce some extra solutions that we removed. Observe that when $\zeta$ tends to $\infty$ the expressions in (\[ahalfsol\]) reduce to (\[aonesol\]). This is in accordance with the fact that $\widetilde{\sigma}$ approaches $\lambda_0$ as $\zeta \to \infty$, see (\[zetadeltabalayage\]), hence we only considered the positive branch of the square root in (\[ahalfsol\]).

Finally take $a=0$. From (\[generalexample\]) we have that $$x_{j,n}\cos \left\{2\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}$$ $$-\sqrt{1-x_{j,n}^2}\sin \left\{2\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}=\cos \kappa_{j,n}.$$ We use the conditions (\[condcos\]) and (\[condsin\]), and obtain the following expression $$x_{j,n}=\frac{2 \zeta \cos \kappa_{j,n}}{\varphi(\zeta) + 