If `0 le x le 2pi`, then `2^(cosec^(2) x) sqrt(1/2 y^(2) -y+1) le sqrt(2)`

Options

A. is satisfied by exactly one value of y
B. is satisfied by exactly two value of x
C. is satisfied by x for which `cos x=0`
D. is satisfied by x for which `sin x=0`

Explanation

To solve the inequality \( 2^{\csc^2 x} \sqrt{\frac{1}{2}y^2 - y + 1} \leq \sqrt{2} \) for \( 0 \leq x \leq 2\pi \), we will break it down step by step.

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ 2^{\csc^2 x} \sqrt{\frac{1}{2}y^2 - y + 1} \leq \sqrt{2} \]
We can square both sides to eliminate the square root (noting that both sides are non-negative):
\[ (2^{\csc^2 x})^2 \left(\frac{1}{2}y^2 - y + 1\right) \leq 2 \]
This simplifies to:
\[ 2^{2\csc^2 x} \left(\frac{1}{2}y^2 - y + 1\right) \leq 2 \]

### Step 2: Isolate the Terms
Next, we can divide both sides by 2 (since 2 is positive):
\[ 2^{2\csc^2 x - 1} \left(\frac{1}{2}y^2 - y + 1\right) \leq 1 \]

### Step 3: Analyze the Terms
Now we analyze the term \( 2^{2\csc^2 x - 1} \). Since \( \csc^2 x = \frac{1}{\sin^2 x} \), we have:
\[ 2^{2\csc^2 x - 1} = 2^{\frac{2}{\sin^2 x} - 1} \]
This term is always greater than or equal to 1 because \( \csc^2 x \geq 1 \) for all \( x \) in the domain.

### Step 4: Set Up Conditions
For the inequality \( 2^{2\csc^2 x - 1} \left(\frac{1}{2}y^2 - y + 1\right) \leq 1 \) to hold, we need:
\[ \frac{1}{2}y^2 - y + 1 \leq \frac{1}{2^{2\csc^2 x - 1}} \]
This means we need to find values of \( y \) such that the left-hand side is less than or equal to the right-hand side.

### Step 5: Solve the Quadratic Inequality
The quadratic \( \frac{1}{2}y^2 - y + 1 \) can be analyzed by finding its vertex and roots. The vertex occurs at:
\[ y = -\frac{b}{2a} = \frac{1}{2 \cdot \frac{1}{2}} = 1 \]
Evaluating at \( y = 1 \):
\[ \frac{1}{2}(1)^2 - (1) + 1 = \frac{1}{2} - 1 + 1 = \frac{1}{2} \]
Since the quadratic opens upwards, the minimum value is \( \frac{1}{2} \) at \( y = 1 \).

### Step 6: Set Conditions on \( y \)
For the inequality to hold:
\[ \frac{1}{2} \leq \frac{1}{2^{2\csc^2 x - 1}} \]
This implies:
\[ 2^{2\csc^2 x - 1} \geq 2 \implies 2\csc^2 x - 1 \geq 1 \implies 2\csc^2 x \geq 2 \implies \csc^2 x \geq 1 \]
This is always true, so we need to find specific values of \( x \).

### Step 7: Find Values of \( x \)
The equality holds when:
\[ 2^{\csc^2 x} = 2 \implies \csc^2 x = 1 \implies \sin x = 1 \text{ or } -1 \]
This occurs at:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 8: Conclusion
Thus, the solution to the inequality is:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \quad \text{and} \quad y = 1 \]