All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin ^2 x-2 \sin x+5}}, \frac{1}{4\sin ^2 y} \leq 1$ also satisfy the equation

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Aicharya · Accepted Answer

$2^{\sqrt{\sin ^2 x-2 \sin x+5}} \cdot 2^{-2 \sin ^2 y} \leq 1$ $\begin{aligned} & 2^{\sqrt{\sin ^2 x-2 \sin x+5}} \leq 2^{2 \sin ^2 y} \ & \sqrt{\sin ^2 x-2 \sin x+5} \leq 2 \sin ^2 y \ & \sqrt{(\sin x-1)^2+4} \leq 2 \sin ^2 y\end{aligned}$ $\begin{aligned} & (\sin x-1)^2+4 \leq 4 \sin ^4 y \ & -1 \leq \sin y \leq 1 \ & 0 \leq \sin ^4 y \leq 1 \ & 	ext { For } \sin y= \pm 1 	ext { (maximum value) } \ & (\sin x-1)^2+4=4\end{aligned}$ $\begin{aligned} & (\sin x-1)^2=0 \ & \sin x=1 \ & \sin x=|\sin y|\end{aligned}$

All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin ^2 x-2 \sin x+5}}, \frac{1}{4\sin ^2 y}…

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