The value of ‘x’ satisfying the equation ${\mathrm{x}}^4-2{\left(\mathrm{xsin}\left(\frac{\mathrm{\pi }}{2}\mathrm{x}\right)\right)}^2+1=0$

${\mathrm{x}}^4+1=2{\mathrm{x}}^2{\sin }^2\left(\frac{\mathrm{\pi }}{2}\mathrm{x}\right)\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=2{\sin }^2\left(\frac{\mathrm{\pi }}{2}\mathrm{x}\right)$
Now, LHS $\ge$ 2 where as RHS $\le$ 2
So, equality holds when $2{\sin }^2\left(\frac{\mathrm{\pi }}{2}\mathrm{x}\right)=2\Rightarrow \mathrm{x}=\pm 1$ 

and $2{\sin }^2\left(\frac{\mathrm{\pi }}{2}\mathrm{x}\right)=2\Rightarrow \mathrm{x}=\pm 1$