If `sin^2theta=(x^2+y^2+1)/(2x)` , then `x` must be `-3` (b) `-2` (c) `1` (d) none of these

Explanation

To solve the equation given by \( \sin^2 \theta = \frac{x^2 + y^2 + 1}{2x} \), we will follow these steps:

### Step 1: Understand the range of \( \sin^2 \theta \)
We know that \( \sin^2 \theta \) is always between 0 and 1, inclusive. Therefore, we can write:
\[ 0 \leq \sin^2 \theta \leq 1 \]

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