The equation ${\sin }^4\mathrm{x}-2{\cos }^2\mathrm{x}+{\mathrm{a}}^2=0$ can be solved if

$$\begin{gathered} {\left({\sin }^2\mathrm{x}\right)}^2-2 {\cos }^2\mathrm{x}+{\mathrm{a}}^2=0 \\ \Rightarrow {\mathrm{a}}^2=2 {\cos }^2\mathrm{x}-{\sin }^4\mathrm{x} \end{gathered}$$

$a^2=2-2{\sin }^2\mathrm{x}-{\sin }^4\mathrm{x}=3-{\left({\mathrm{Sin}}^2\mathrm{x}+1\right)}^2$

Now $0\le {\sin }^2\mathrm{x}\le 1 \therefore 1\le {\sin }^2\mathrm{x}+1\le 2 \therefore 1\le {\left({\sin }^2\mathrm{x}+1\right)}^2⩽4$

$$\begin{gathered} \therefore -4\le -{\left({\sin }^2\mathrm{x}+1\right)}^2\le -1 \\ \Rightarrow -1\le 3-{\left({\sin }^2\mathrm{x}+1\right)}^2\le 2\Rightarrow a^2\le 2 \end{gathered}$$

$\therefore {\mathrm{a}}^2\le 2\Rightarrow -\sqrt{2}\le \mathrm{a}\le \sqrt{2}$