All the pairs (x, y) that satisfy the inequal | vedclass

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(D) The given inequality is $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 2^{-2 \sin^2 y} \leq 1$. This simplifies to $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq

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$\sin x = |\sin y|$

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(D) The given inequality is $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 2^{-2 \sin^2 y} \leq 1$. This simplifies to $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 2^{2 \sin^2 y}$. Since the base $2 > 1$,we have $\sqrt{(\sin x - 1)^2 + 4} \leq 2 \sin^2 y$. We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$. Thus,$2 \sin^2 y \geq 2$,which implies $\sin^2 y \geq 1$. Since the maximum value of $\sin^2 y$ is $1$,we must have $\sin^2 y = 1$,which means $\sin y = \pm 1$. Substituting $\sin^2 y = 1$ into the inequality,we get $\sqrt{(\sin x - 1)^2 + 4} \leq 2(1) = 2$. This implies $(\sin x - 1)^2 + 4 \leq 4$,so $(\sin x - 1)^2 \leq 0$. Since a square cannot be negative,we must have $(\sin x - 1)^2 = 0$,which means $\sin x = 1$. Since $\sin x = 1$ and $|\sin y| = 1$,we conclude $\sin x = |\sin y|$.

All the pairs $(x, y)$ 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

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