All the pairs $(\mathrm{x},\mathrm{y})$ that satisfy the inequality 

$\frac{2\sqrt{{\sin }^2\mathrm{x}-2\mathrm{sinx}+5}}{4 {\sin }^2 \mathrm{y}}\le 1$ 

also satisfy the equation:

$\frac{2\sqrt{{\sin }^2\mathrm{x}-2\mathrm{sinx}+5}}{4 {\sin }^2 \mathrm{y}}\le 1$

$$\begin{gathered} \sqrt{{\sin }^2\mathrm{x}-2\mathrm{sinx}+5}\le 2{\sin }^2\mathrm{y} \\ {\sin }^2\mathrm{x}-2\mathrm{sinx}+5\le 4{\sin }^4\mathrm{y} \\ {(\sin \mathrm{x}-1)}^2+4\le 4 {\sin }^4\mathrm{y} \\ \sin \mathrm{x}=1, \sin \mathrm{y}=\pm 1 \\ \sin \mathrm{x}=\left|\sin \mathrm{y}\right| \end{gathered}$$