Find the total no. of orderd pairs (x,y) satisfying `x(sin^2 x+ 1/x^2)=2sinxsin^2y,` where `x in (-pi,0)uu(0,pi) and y in[0,2pi]` .

Explanation

To solve the equation \( x(\sin^2 x + \frac{1}{x^2}) = 2 \sin x \sin^2 y \) for the ordered pairs \( (x, y) \) satisfying the given conditions, we will follow these steps:

### Step 1: Analyze the given equation
We start with the equation:
\[ x(\sin^2 x + \frac{1}{x^2}) = 2 \sin x \sin^2 y \]
We need to analyze the left-hand side and the right-hand side separately.

### Step 2: Simplify the left-hand side
The left-hand side can be rewritten as:
\[ x \sin^2 x + \frac{x}{x^2} = x \sin^2 x + \frac{1}{x} \]
This means we need to analyze the behavior of \( x \sin^2 x + \frac{1}{x} \).

### Step 3: Find the range of \( x \)
The variable \( x \) is defined in the intervals \( (-\pi, 0) \cup (0, \pi) \). We will consider both intervals separately.

### Step 4: Analyze \( \sin^2 x + \frac{1}{x^2} \)
Using the AM-GM inequality:
\[ \frac{\sin^2 x + \frac{1}{x^2}}{2} \geq \sqrt{\sin^2 x \cdot \frac{1}{x^2}} \]
This implies:
\[ \sin^2 x + \frac{1}{x^2} \geq \frac{2 \sin x}{|x|} \]
Thus, we can express the left-hand side as:
\[ x \left( \sin^2 x + \frac{1}{x^2} \right) \geq 2 \sin x \]

### Step 5: Set up the inequality
From the previous analysis, we conclude:
\[ x \left( \sin^2 x + \frac{1}{x^2} \right) \geq 2 \sin x \]
This means that the right-hand side \( 2 \sin x \sin^2 y \) must also be less than or equal to \( 2 \sin x \).

### Step 6: Analyze \( \sin^2 y \)
Since \( \sin^2 y \) must be less than or equal to 1, we have:
\[ \sin^2 y \leq 1 \]
This means that \( \sin^2 y = 1 \) is the only case that satisfies the equality condition.

### Step 7: Solve for \( y \)
The equation \( \sin^2 y = 1 \) implies:
\[ y = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Within the interval \( [0, 2\pi] \), the valid solutions for \( y \) are:
\[ y = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 8: Solve for \( x \)
Now we need to find \( x \) such that:
\[ x \sin^2 x + \frac{1}{x} = 2 \sin x \]
This will yield two intersection points for \( x \) in the intervals \( (-\pi, 0) \) and \( (0, \pi) \).

### Step 9: Count the solutions
From our analysis, we find:
- There are 2 valid \( x \) values in \( (-\pi, 0) \) and \( (0, \pi) \).
- For each \( x \), there are 2 valid \( y \) values.

Thus, the total number of ordered pairs \( (x, y) \) is:
\[ 2 \text{ (values of } x) \times 2 \text{ (values of } y) = 4 \]

### Final Answer
The total number of ordered pairs \( (x, y) \) satisfying the given equation is **4**.

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