The number of ordered pairs which satisfy the equation `x^2+2xsin(x y)+1=0` are (where `y in [0,2pi]` ) (a) 1 (b) 2 (c) 3 (d) 0

Explanation

To solve the equation \( x^2 + 2x \sin(xy) + 1 = 0 \) for ordered pairs \((x, y)\) where \( y \in [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ x^2 + 2x \sin(xy) + 1 = 0 \]
This is a quadratic equation in terms of \( x \).
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