Question:
All the pairs $(\mathrm{x}, \mathrm{y})$ that satisfy the inequality $2 \sqrt{\sin ^{2} x-2 \sin x+5} \cdot \frac{1}{4 \sin ^{2} y} \leq 1$ also satisfy the equation:
Correct Option: , 4
Solution:
Given inequality is,
$2 \sqrt{\sin ^{2} x-2 \sin x+5} \leq 2^{2 \sin ^{2} y}$
$\Rightarrow \sqrt{\sin ^{2} x-2 \sin x+5} \leq 2 \sin ^{2} y$
$\Rightarrow \sqrt{(\sin x-1)^{2}+4} \leq 2 \sin ^{2} y$
It is true if $\sin x=1$ and $|\sin y|=1$
Therefore, $\sin x=|\sin y|$