All the pairs `(x, y)` that satisfy the inequality
`2 ^(sqrt(sin^(2) x - 2 sin x + 5) ) . (1)/(4 ^(sin^(2) y)) le 1` also satisfy the equation :

Options

A. `12/5sqrt(2)`
B. `11/6sqrt(2)`
C. `12/5sqrt(3)`
D. `11/6sqrt(3)`

Answer

To solve the inequality \( 2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \leq 1 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the inequality in a more manageable form:
\[ 2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 4^{-\sin^2 y} \leq 1 \]
Since \( 4 = 2^2 \), we can rewrite \( 4^{-\sin^2 y} \) as \( 2^{-2 \sin^2 y} \):
\[ 2^{\sqrt{\sin^2 x - 2 \sin x + 5} - 2 \sin^2 y} \leq 1 \]

### Step 2: Analyze the exponent
The inequality \( 2^a \leq 1 \) holds when \( a \leq 0 \). Thus, we can set up the following inequality:
\[ \sqrt{\sin^2 x - 2 \sin x + 5} - 2 \sin^2 y \leq 0 \]

### Step 3: Rearranging the inequality
Rearranging gives us:
\[ \sqrt{\sin^2 x - 2 \sin x + 5} \leq 2 \sin^2 y \]

### Step 4: Square both sides
To eliminate the square root, we square both sides (noting that both sides are non-negative):
\[ \sin^2 x - 2 \sin x + 5 \leq 4 \sin^4 y \]

### Step 5: Rearranging the terms
Rearranging the terms gives us:
\[ \sin^2 x - 2 \sin x + 5 - 4 \sin^4 y \leq 0 \]

### Step 6: Finding the range of \( \sqrt{\sin^2 x - 2 \sin x + 5} \)
Next, we need to analyze the expression \( \sin^2 x - 2 \sin x + 5 \):
\[ \sin^2 x - 2 \sin x + 5 = (\sin x - 1)^2 + 4 \]
This expression is always greater than or equal to 4, since the minimum value of \((\sin x - 1)^2\) is 0 (when \(\sin x = 1\)).

### Step 7: Establishing the range of \( 2 \sin^2 y \)
The expression \( 2 \sin^2 y \) ranges from 0 to 2, since \(\sin^2 y\) ranges from 0 to 1.

### Step 8: Setting up the equality condition
The inequality holds if and only if:
\[ (\sin x - 1)^2 + 4 = 2 \sin^2 y \]
This equality can only occur when both sides are equal to 2, which leads us to the conditions:
\[ \sin x = 1 \quad \text{and} \quad \sin^2 y = 1 \]

### Step 9: Final conditions
Thus, we find:
\[ \sin x = 1 \implies x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
\[ \sin y = \pm 1 \implies y = \frac{\pi}{2} + m\pi \quad (m \in \mathbb{Z}) \]

### Conclusion
The pairs \((x, y)\) that satisfy the inequality are:
\[ (x, y) = \left(\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + m\pi\right) \quad (n, m \in \mathbb{Z}) \]