The number of solutions of the equation $2{\left({\sin }^{-1}x\right)}^2-5{\sin }^{-1}x+2=0$  is

$\Rightarrow 2{\left({\sin }^{-1}x\right)}^2-4{\sin }^{-1}x-{\sin }^{-1}x+2=0$

$\Rightarrow 2{\sin }^{-1}x\left[{\sin }^{-1}x-2\right]-1\left({\sin }^{-1}x-2\right)=0$

$\Rightarrow \left(2{\sin }^{-1}x-1\right)\left({\sin }^{-1}x-2\right)=0$

$\Rightarrow {\sin }^{-1}x=\frac{1}{2}$  (or) ${\sin }^{-1}x=2$

$x=\sin \left(\frac{1}{2}\right)\left(or\right)\sin \left(2\right)$