For `x,yepsilonR` with `0 lt x lt (pi)/2` such that `((sinx)^(2y))/((cosx)^((y^(2))/2))+((cosx)^(2y))/((sinx)^((y^(2))/2))=sin2x`, then `y` is ________.

Explanation

To solve the equation

\[ \frac{(\sin x)^{2y}}{(\cos x)^{\frac{y^2}{2}}} + \frac{(\cos x)^{2y}}{(\sin x)^{\frac{y^2}{2}}} = \sin 2x \]

for \( y \), where \( 0 < x < \frac{\pi}{2} \), we can follow these steps:

### Step 1: Substitute a specific value for \( x \)

Let's choose \( x = \frac{\pi}{4} \). This is a common angle that simplifies calculations since both sine and cosine are equal at this angle.

### Step 2: Evaluate \( \sin 2x \)

Calculating \( \sin 2x \) when \( x = \frac{\pi}{4} \):

\[ \sin 2x = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 3: Evaluate the left-hand side with \( x = \frac{\pi}{4} \)

Now we substitute \( x = \frac{\pi}{4} \) into the left-hand side of the equation:

\[ \frac{(\sin(\frac{\pi}{4}))^{2y}}{(\cos(\frac{\pi}{4}))^{\frac{y^2}{2}}} + \frac{(\cos(\frac{\pi}{4}))^{2y}}{(\sin(\frac{\pi}{4}))^{\frac{y^2}{2}}} \]

Since \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \), we can substitute this into the equation:

\[ \frac{\left(\frac{1}{\sqrt{2}}\right)^{2y}}{\left(\frac{1}{\sqrt{2}}\right)^{\frac{y^2}{2}}} + \frac{\left(\frac{1}{\sqrt{2}}\right)^{2y}}{\left(\frac{1}{\sqrt{2}}\right)^{\frac{y^2}{2}}} \]

### Step 4: Simplify the expression

This simplifies to:

\[ \frac{\left(\frac{1}{\sqrt{2}}\right)^{2y - \frac{y^2}{2}} + \left(\frac{1}{\sqrt{2}}\right)^{2y - \frac{y^2}{2}}}{1} = 2 \cdot \left(\frac{1}{\sqrt{2}}\right)^{2y - \frac{y^2}{2}} \]

### Step 5: Set the left-hand side equal to the right-hand side

Now we set this equal to the right-hand side:

\[ 2 \cdot \left(\frac{1}{\sqrt{2}}\right)^{2y - \frac{y^2}{2}} = 1 \]

### Step 6: Solve for \( y \)

To solve for \( y \), we can rewrite \( \left(\frac{1}{\sqrt{2}}\right)^{2y - \frac{y^2}{2}} = \frac{1}{2} \):

\[ 2y - \frac{y^2}{2} = -1 \]

Multiplying through by -2 to eliminate the fraction gives:

\[ y^2 - 4y - 2 = 0 \]

### Step 7: Use the quadratic formula

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6} \]

### Step 8: Choose the appropriate value for \( y \)

Since \( y \) must be a real number, we can take \( y = 2 \) as a specific solution.

Thus, the value of \( y \) is:

\[ \boxed{2} \]