the number of solution of the equation `1+ sinx.sin^2(x/2) = 0` , in `[-pi , pi ]` , is

Options

A. 0
B. 1
C. 3
D. none of these

Explanation

To solve the equation \(1 + \sin x \cdot \sin^2\left(\frac{x}{2}\right) = 0\) in the interval \([- \pi, \pi]\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 1 + \sin x \cdot \sin^2\left(\frac{x}{2}\right) = 0 \]
This can be rearranged to:
\[ \sin x \cdot \sin^2\left(\frac{x}{2}\right) = -1 \]

### Step 2: Analyzing the Range of the Left Side
Next, we analyze the left side of the equation, \( \sin x \cdot \sin^2\left(\frac{x}{2}\right) \).

1. The function \( \sin x \) varies between \(-1\) and \(1\) for \(x \in [-\pi, \pi]\).
2. The function \( \sin^2\left(\frac{x}{2}\right) \) varies between \(0\) and \(1\) since squaring a sine function always yields a non-negative result.

### Step 3: Finding the Range of the Product
Now, we consider the product:
\[ \sin x \cdot \sin^2\left(\frac{x}{2}\right) \]
- The maximum value of \( \sin x \) is \(1\) and the minimum is \(-1\).
- The maximum value of \( \sin^2\left(\frac{x}{2}\right) \) is \(1\) (when \(x = 0\)).
- Therefore, the product \( \sin x \cdot \sin^2\left(\frac{x}{2}\right) \) will range from \(0\) (when \( \sin x = 0 \) or \( \sin^2\left(\frac{x}{2}\right) = 0 \)) to \(1\) (when \( \sin x = 1\) and \( \sin^2\left(\frac{x}{2}\right) = 1\)).

### Step 4: Conclusion on the Range
Since the left-hand side \( \sin x \cdot \sin^2\left(\frac{x}{2}\right) \) can only take values in the range \([-1, 1]\), it cannot equal \(-1\) because the product is always non-negative or less than \(1\).

### Step 5: Final Result
Thus, the equation \(1 + \sin x \cdot \sin^2\left(\frac{x}{2}\right) = 0\) has no solutions in the interval \([- \pi, \pi]\).

### Answer
The number of solutions of the equation in the interval \([- \pi, \pi]\) is:
\[ \boxed{0} \]