If $x\left(3-x\right)\ge 2$ then ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(x^2\right)+\mathrm{....}+{\sin }^2\left(x^{10}\right)=$

$$\begin{gathered} Given x\left(3-x\right)\ge 2\Rightarrow x^2-3x+2\le 0\Rightarrow \left(x-1\right)\left(x-2\right)\le 0\Rightarrow 1\le x\le 2 \\ But {\sin }^{-1}x is defined if -1\le x\le 1 \\ \Rightarrow x=1 \\ \therefore {\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(x^2\right)+.........+{\sin }^{-1}\left(x^{10}\right)={\sin }^{-1}\left(1\right)+{\sin }^{-1}\left(1\right)+.....+{\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}+......+\frac{\mathrm{\pi }}{2} \\ =5\mathrm{\pi } \\ =5\times 3.14 \\ =15.70 \end{gathered}$$