Suppose that for some angles `xa n dy ,` the equations `sin^2x+cos^2y=(3a)/2a n dcos^2x+sin^2y=(a^2)/2` hold simultaneously. the possible value of `a` is ___________

Explanation

To solve the problem, we need to analyze the given equations step by step.

### Given Equations:
1. \( \sin^2 x + \cos^2 y = \frac{3a}{2} \) (Equation 1)
2. \( \cos^2 x + \sin^2 y = \frac{a^2}{2} \) (Equation 2)

### Step 1: Add the two equations
We start by adding Equation 1 and Equation 2:

\[ (\sin^2 x + \cos^2 y) + (\cos^2 x + \sin^2 y) = \frac{3a}{2} + \frac{a^2}{2} \]

### Step 2: Simplify the left-hand side
Using the Pythagorean identity, we know that:

\[ \sin^2 x + \cos^2 x = 1 \quad \text{and} \quad \sin^2 y + \cos^2 y = 1 \]

Thus, we can rewrite the left-hand side:

\[ 1 + 1 = \frac{3a}{2} + \frac{a^2}{2} \]

This simplifies to:

\[ 2 = \frac{3a + a^2}{2} \]

### Step 3: Eliminate the fraction
Multiply both sides by 2 to eliminate the fraction:

\[ 4 = 3a + a^2 \]

### Step 4: Rearrange the equation
Rearranging gives us a standard quadratic equation:

\[ a^2 + 3a - 4 = 0 \]

### Step 5: Factor the quadratic equation
Now we will factor the quadratic equation:

\[ (a + 4)(a - 1) = 0 \]

### Step 6: Solve for \(a\)
Setting each factor to zero gives us the possible solutions:

1. \( a + 4 = 0 \) → \( a = -4 \)
2. \( a - 1 = 0 \) → \( a = 1 \)

### Step 7: Check the validity of the solutions
We need to check which of these values satisfies the original equations.

#### Checking \( a = -4 \):
Substituting \( a = -4 \) into Equation 1:

\[ \sin^2 x + \cos^2 y = \frac{3(-4)}{2} = -6 \]

This is not possible since \( \sin^2 x + \cos^2 y \) must be non-negative.

#### Checking \( a = 1 \):
Substituting \( a = 1 \) into Equation 1:

\[ \sin^2 x + \cos^2 y = \frac{3(1)}{2} = \frac{3}{2} \]

This is valid since \( 0 \leq \sin^2 x + \cos^2 y \leq 2 \).

### Conclusion
The only valid solution is:

\[ \boxed{1} \]

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