If `sin^(2)theta=(x^(2)+y^(2)+1)/(2x)`. Find the value of x and y.

Answer

To solve the equation \( \sin^2 \theta = \frac{x^2 + y^2 + 1}{2x} \) and find the values of \( x \) and \( y \), we can follow these steps:

### Step 1: Understand the Range of \( \sin^2 \theta \)
The value of \( \sin^2 \theta \) lies between 0 and 1, inclusive. Therefore, we can write:
\[ 0 \leq \sin^2 \theta \leq 1 \]

### Step 2: Set Up the Inequality
From the equation given, we can express this as:
\[ 0 \leq \frac{x^2 + y^2 + 1}{2x} \leq 1 \]
This implies:
\[ 0 \leq x^2 + y^2 + 1 \leq 2x \]

### Step 3: Rearrange the Inequality
We can rearrange the inequality \( x^2 + y^2 + 1 \leq 2x \):
\[ x^2 - 2x + y^2 + 1 \leq 0 \]

### Step 4: Complete the Square
Now, we can complete the square for the \( x \) terms:
\[ (x - 1)^2 + y^2 \leq 0 \]

### Step 5: Analyze the Result
Since both \( (x - 1)^2 \) and \( y^2 \) are squares, they are always non-negative. The only way their sum can be less than or equal to zero is if both terms are exactly zero:
\[ (x - 1)^2 = 0 \quad \text{and} \quad y^2 = 0 \]

### Step 6: Solve for \( x \) and \( y \)
From \( (x - 1)^2 = 0 \), we get:
\[ x - 1 = 0 \implies x = 1 \]
From \( y^2 = 0 \), we get:
\[ y = 0 \]

### Conclusion
Thus, the values of \( x \) and \( y \) are:
\[ \boxed{x = 1, y = 0} \]