Step 1: Rewrite the inequality.
Given: $2^{\sqrt{\sin^2 x-2\sin x+5}} - \frac{1}{4^{\sin y}} \leq 1$. Note $\frac{1}{4^{\sin y}} = 2^{-2\sin y}$, so the inequality is: \[ 2^{\sqrt{\sin^2 x-2\sin x+5}} - 2^{-2\sin y} \leq 1 \]
Step 2: Find the minimum of the exponent in the first term.
$\sin^2 x - 2\sin x + 5 = (\sin x - 1)^2 + 4 \geq 4$. So $\sqrt{(\sin x-1)^2+4} \geq 2$, meaning $2^{\sqrt{...}} \geq 4$.
Step 3: Find the range of the second term.
Since $-1 \leq \sin y \leq 1$, we have $-2 \leq -2\sin y \leq 2$, so $2^{-2\sin y} \in [2^{-2}, 2^2] = [\frac{1}{4}, 4]$. The maximum is $4$ (when $\sin y = -1$).
Step 4: Determine when the inequality holds.
The left side is at least $4 - 4 = 0$. For it to be $\leq 1$, we need both: the first term to be at its minimum $4$, and the second term to be at its maximum $4$. So both conditions must hold simultaneously: $\sin x = 1$ AND $\sin y = -1$.
Step 5: Identify the required equation.
When $\sin x = 1$ and $\sin y = -1$: $|\sin y| = 1 = \sin x$. So $\sin x = |\sin y|$. Check other options: $2|\sin x| = 2(1)=2 \neq -1=\sin y$ (option 1 fails).
Step 6: State the answer.
\[ \boxed{\sin x = |\sin y|} \]