`sin^(2)theta=(x^(2)+y^(2)+1)/(2x)`, then x must be.

Answer

To solve the equation \( \sin^2 \theta = \frac{x^2 + y^2 + 1}{2x} \), we need to analyze the constraints on \( x \) based on the properties of the sine function.

### Step-by-Step Solution:

1. **Understanding the Range of \( \sin^2 \theta \)**:
\[ 0 \leq \sin^2 \theta \leq 1 \]
Since \( \sin \theta \) can take values between -1 and 1, squaring it gives us a range from 0 to 1.

**Hint**: Remember that squaring a number between -1 and 1 results in a value between 0 and 1.

2. **Setting Up the Inequality**:
From the equation, we have:
\[ \frac{x^2 + y^2 + 1}{2x} \leq 1 \]
Multiplying both sides by \( 2x \) (assuming \( x > 0 \) to avoid division by zero):
\[ x^2 + y^2 + 1 \leq 2x \]
Rearranging gives:
\[ x^2 - 2x + y^2 + 1 \leq 0 \]

**Hint**: When rearranging inequalities, ensure to maintain the direction of the inequality.

3. **Completing the Square**:
The expression \( x^2 - 2x + 1 \) can be rewritten as:
\[ (x - 1)^2 \]
Thus, the inequality becomes:
\[ (x - 1)^2 + y^2 \leq 0 \]

**Hint**: Completing the square helps in identifying the nature of the quadratic expression.

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

**Hint**: A sum of non-negative numbers can only be zero if each number is zero.

5. **Solving for \( x \) and \( y \)**:
From \( (x - 1)^2 = 0 \):
\[ x - 1 = 0 \implies x = 1 \]
From \( y^2 = 0 \):
\[ y = 0 \]

**Hint**: Solving a squared equation gives you the roots directly.

### Conclusion:
Thus, the value of \( x \) must be:
\[ \boxed{1} \]