`x^(2)-2x+2cos^(2)theta +sin^(2)theta=0`, then maximum number of ordered pair `(x, theta)` such that `x ∈ R, theta ∈ [0,2pi]`.

Answer

To solve the equation \( x^2 - 2x + 2\cos^2\theta + \sin^2\theta = 0 \) and find the maximum number of ordered pairs \( (x, \theta) \) such that \( x \in \mathbb{R} \) and \( \theta \in [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ x^2 - 2x + 2\cos^2\theta + \sin^2\theta = 0 \]
We know that \( \sin^2\theta + \cos^2\theta = 1 \), so we can substitute \( \sin^2\theta = 1 - \cos^2\theta \):
\[ x^2 - 2x + 2\cos^2\theta + (1 - \cos^2\theta) = 0 \]
This simplifies to:
\[ x^2 - 2x + (1 + \cos^2\theta) = 0 \]

### Step 2: Analyze the quadratic equation
The equation \( x^2 - 2x + (1 + \cos^2\theta) = 0 \) is a quadratic in \( x \). For this quadratic to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(1 + \cos^2\theta) \geq 0 \]
Calculating the discriminant:
\[ D = 4 - 4(1 + \cos^2\theta) = 4 - 4 - 4\cos^2\theta = -4\cos^2\theta \]
For \( D \geq 0 \), we need:
\[ -4\cos^2\theta \geq 0 \]
This implies:
\[ \cos^2\theta = 0 \]

### Step 3: Solve for \( \theta \)
The condition \( \cos^2\theta = 0 \) means:
\[ \cos\theta = 0 \]
The values of \( \theta \) that satisfy this in the interval \( [0, 2\pi] \) are:
\[ \theta = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 4: Find corresponding \( x \) values
Substituting \( \cos^2\theta = 0 \) back into the quadratic equation:
\[ x^2 - 2x + 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 = 0 \]
Thus, the only solution for \( x \) is:
\[ x = 1 \]

### Step 5: Determine the ordered pairs
The ordered pairs \( (x, \theta) \) that satisfy the conditions are:
1. \( (1, \frac{\pi}{2}) \)
2. \( (1, \frac{3\pi}{2}) \)

### Conclusion
The maximum number of ordered pairs \( (x, \theta) \) such that \( x \in \mathbb{R} \) and \( \theta \in [0, 2\pi] \) is:
\[ \boxed{2} \]