Let $(\mathrm{a},\mathrm{b})\subset (0,2\mathrm{\pi })$ be the largest interval for which ${\sin }^{-1}(\sin \mathrm{\theta })-{\cos }^{-1}(\sin \mathrm{\theta })>0,\mathrm{\theta }\in (0,2\mathrm{\pi })$, holds 
If ${\mathrm{\alpha x}}^2+\mathrm{\beta x}+{\sin }^{-1}\left({\mathrm{x}}^2-6\mathrm{x}+10\right)+{\cos }^{-1}\left({\mathrm{x}}^2-6\mathrm{x}+10\right)=0\text{ and }\mathrm{\alpha }-\mathrm{\beta }=\mathrm{b}-\mathrm{a},\text{ the }\mathrm{\alpha }$ is equal to :

$\begin{array}{l}{\sin }^{-1}(\sin \mathrm{\theta })-{\cos }^{-1}(\sin \mathrm{\theta })>0\text{ for }\mathrm{\theta }\in (0,2\mathrm{\pi })\\ {\sin }^{-1}(\sin \mathrm{\theta })-\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}(\sin \mathrm{\theta })\right)>0\\ \Rightarrow {\sin }^{-1}(\sin \mathrm{\theta })>\frac{\mathrm{\pi }}{4}\\ \frac{1}{\sqrt{2}}<\sin \mathrm{\theta }\le 1\Rightarrow \frac{\mathrm{\pi }}{4}<\mathrm{\theta }<\frac{3\mathrm{\pi }}{4}\\ \mathrm{a}=\frac{\mathrm{\pi }}{4},\mathrm{b}=\frac{3\mathrm{\pi }}{4},\mathrm{b}-\mathrm{a}=\frac{\mathrm{\pi }}{2}\end{array}$
${\mathrm{\alpha x}}^2+\mathrm{\beta x}+{\sin }^{-1}\left((\mathrm{x}-3{)}^2+1\right)+\left({\cos }^{-1}(\mathrm{x}-3{)}^2+1\right)=0$
is true only for x = 3
$\begin{array}{l}9\mathrm{\alpha }+3\mathrm{\beta }+\frac{\mathrm{\pi }}{2}+0=0\Rightarrow 3\mathrm{\alpha }+\mathrm{\beta }=\frac{-\mathrm{\pi }}{6}\dots .\left(1\right)\\ \mathrm{\alpha }-\mathrm{\beta }=\mathrm{b}-\mathrm{a}\Rightarrow \mathrm{\alpha }-\mathrm{\beta }=\frac{\mathrm{\pi }}{2} ...\left(2\right)\end{array}$
 $4\mathrm{\alpha }=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3}\Rightarrow \mathrm{\alpha }=\frac{\mathrm{\pi }}{12}$