If $0\le \mathrm{x}\le 2\mathrm{\pi }$ and $2^{{\mathrm{cosec}}^{2}x}\sqrt{\left(\frac{1}{2}{\mathrm{y}}^2-\mathrm{y}+1\right)}\le \sqrt{2}$ then number of ordered pairs of (x,y) is

$2^{{\mathrm{cosec}}^2\mathrm{x}}\sqrt{\left(\frac{1}{2}{\mathrm{y}}^2-\mathrm{y}+1\right)}\le \sqrt{2}$

$\Rightarrow 2^{{\mathrm{cosec}}^2\mathrm{x}}\sqrt{\left.{\mathrm{y}}^2-2\mathrm{y}+2\right)}\le 2$

$2^{{\mathrm{cosec}}^2\mathrm{x}}\sqrt{\left\{(\mathrm{y}-1{)}^2+1\right\}}\le 2----\left(\mathrm{i}\right)$

We have

$\therefore 2^{{\text{cosec }}^2\mathrm{x}}\ge 2\text{ and }\Rightarrow \sqrt{(\mathrm{y}-1{)}^2+1}\ge 1$

$\Rightarrow \text{ product }\ge 2----\left(\mathrm{ii}\right)$

From (i) and (ii)

$\begin{array}{r}{2}^{\mathrm{cosec}{\mathrm{x}}^2\mathrm{x}}=2\text{ and }\sqrt{\left\{(\mathrm{y}-1{)}^2+1\right\}}=1\\ \Rightarrow \sin \mathrm{x}=\pm 1\text{ and }\mathrm{y}=1\\ \Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\text{ and }\mathrm{y}=1\end{array}$

Hence, the solution of the given inequality is $(\mathrm{x},\mathrm{y})=\left(\frac{\mathrm{\pi }}{2},1\right),\left(\frac{3\mathrm{\pi }}{2},1\right)$