Find all number x , y that satisfy the equation `(sin^2 x +(1)/( sin^(2) x))^(2)+( cos^(2)x+(1)/(cos^(2)x))^(2)=12+(1)/(2) siny. `

Answer

To solve the equation

\[ \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 = 12 + \frac{1}{2} \sin y, \]

we will follow these steps:

### Step 1: Use the AM-GM Inequality
We know that for any positive real numbers \(a\) and \(b\):

\[ \frac{a + b}{2} \geq \sqrt{ab}. \]

Applying this to \(\sin^2 x\) and \(\frac{1}{\sin^2 x}\):

\[ \sin^2 x + \frac{1}{\sin^2 x} \geq 2. \]

Similarly, for \(\cos^2 x\):

\[ \cos^2 x + \frac{1}{\cos^2 x} \geq 2. \]

### Step 2: Square the Inequalities
Now squaring both inequalities:

\[ \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 \geq 4, \]
\[ \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 \geq 4. \]

### Step 3: Combine the Results
Adding these inequalities gives:

\[ \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 \geq 4 + 4 = 8. \]

### Step 4: Establish the Upper Bound
Now, we can find the maximum value of \(\sin^2 x + \frac{1}{\sin^2 x}\) and \(\cos^2 x + \frac{1}{\cos^2 x}\). The maximum occurs when \(\sin^2 x = 1\) or \(\cos^2 x = 1\):

\[ \sin^2 x + \frac{1}{\sin^2 x} \leq 2 + 2 = 4, \]
\[ \cos^2 x + \frac{1}{\cos^2 x} \leq 2 + 2 = 4. \]

Thus:

\[ \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 \leq 16. \]

### Step 5: Set Up the Inequality
From the original equation, we have:

\[ 8 \leq 12 + \frac{1}{2} \sin y \leq 16. \]

### Step 6: Solve for \(\sin y\)
From \(8 \leq 12 + \frac{1}{2} \sin y\):

\[ -4 \leq \frac{1}{2} \sin y \implies -8 \leq \sin y. \]

Since \(\sin y\) is always between -1 and 1, this condition is always satisfied.

From \(12 + \frac{1}{2} \sin y \leq 16\):

\[ \frac{1}{2} \sin y \leq 4 \implies \sin y \leq 8. \]

This is also always satisfied since \(\sin y\) is bounded by -1 and 1.

### Step 7: Find Values of \(x\)
Now, we need to find values of \(x\) such that:

\[ \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 = 12 + \frac{1}{2} \sin y. \]

The maximum value of the left-hand side is 16, which occurs when both \(\sin^2 x\) and \(\cos^2 x\) are 1. This gives us:

\[ 12 + \frac{1}{2} \sin y = 16 \implies \frac{1}{2} \sin y = 4 \implies \sin y = 8. \]

This is impossible, so we check the minimum:

\[ 12 + \frac{1}{2} \sin y = 8 \implies \frac{1}{2} \sin y = -4 \implies \sin y = -8. \]

This is also impossible.

### Conclusion
Thus, the only valid solutions for \(x\) occur when:

\[ \sin^2 x = \frac{1}{2} \implies x = n\pi + \frac{\pi}{4}, n \in \mathbb{Z}, \]
and for \(y\):

\[ \sin y = 1 \implies y = \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}. \]

### Final Answer
The solutions are:

\[ x = n\pi + \frac{\pi}{4}, \quad y = \frac{\pi}{2} + 2m\pi, \quad n, m \in \mathbb{Z}. \]