If the equation  ${\left(2^{\left(\frac{1}{{\cos }^{-1}x}\right)}\right)}^{2\pi }-{\left(2^{\left(\frac{1}{{\cos }^{-1}x}\right)}\right)}^{\pi }\left(a+\frac{1}{2}\right)-a^2=0$   has only one real solution then subsets of values of ‘a’ are

$\begin{array}{l}let 2^{\left(\frac{\pi }{{\cos }^{-1}\text{x }}\right)}=t \\ \text{ since }0\le {\cos }^{-1}\text{x}\le \pi \Rightarrow \frac{1}{\pi }\le \frac{1}{{\cos }^{-1}\text{x}}\le \infty \Rightarrow 1\le \frac{\pi }{{\cos }^{-1}\text{x}}\le \infty \Rightarrow 2\le t<\infty \to \left(1\right)\\ 2^{2\pi /{\cos }^{-1}\text{x}}-\left(\text{a}+\frac{1}{2}\right)2^{\pi /{\cos }^{-1}\text{x}}-{\text{a}}^2=0\end{array}$

$\begin{array}{l}\Rightarrow {\text{t}}^2-\left(\text{a}+\frac{1}{2}\right)\text{t}-{\text{a}}^2=0\Rightarrow \text{t}=\frac{\left(\text{a}+\frac{1}{2}\right)\pm \sqrt{{\left(\text{a}+\frac{1}{2}\right)}^2+4{\text{a}}^2}}{2}\\ \text{ From (1) }\\ 2\le \frac{\left(a+\frac{1}{2}\right)\pm \sqrt{{\left(a+\frac{1}{2}\right)}^2+4a^2}}{2}\le \infty \\ \Rightarrow 4\le \left(a+\frac{1}{2}\right)\pm \sqrt{{\left(a+\frac{1}{2}\right)}^2+4a^2}\le \infty \\ \Rightarrow \pm \sqrt{{\left(a+\frac{1}{2}\right)}^2+4a^2}\ge 4-\left(a+\frac{1}{2}\right)=\frac{7}{2}-a\\ \Rightarrow {\left(a+\frac{1}{2}\right)}^2+4a^2\ge {\left(\frac{7}{2}-a\right)}^2\end{array}$

$\begin{array}{c}\Rightarrow {\text{a}}^2+\frac{1}{4}+\text{a}+4{\text{a}}^2\ge \frac{49}{4}+{\text{a}}^2-7\text{a}\\ \Rightarrow 4{\text{a}}^2+8\text{a}-12\ge 0\Rightarrow {\text{a}}^2+2\text{a}-3\ge 0\\ \Rightarrow (\text{a}-1)(\text{a}+3)\ge 0\Rightarrow \text{a}\le -3\text{ or a}\ge 1\end{array}$