The number of solutionsof the equation
`1 +sin x sin^2 ""(x)/2 = 0` in `[-pi,pi]` is

Options

A. zero
B. 1
C. 2
D. 3

Answer

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

### Step 1: Rearranging the equation
We start with the equation:
\[ 1 + \sin x \sin^2 \left( \frac{x}{2} \right) = 0 \]
Rearranging gives us:
\[ \sin x \sin^2 \left( \frac{x}{2} \right) = -1 \]

### Step 2: Analyzing the range of sine functions
We know that the sine function, \( \sin x \), has a range of \([-1, 1]\) and \( \sin^2 \left( \frac{x}{2} \right) \) has a range of \([0, 1]\). Therefore, the product \( \sin x \sin^2 \left( \frac{x}{2} \right) \) will also have a range of \([-1, 1]\).

### Step 3: Finding the impossibility of the equation
The equation \( \sin x \sin^2 \left( \frac{x}{2} \right) = -1 \) implies that the left side must equal \(-1\). However, since \( \sin x \sin^2 \left( \frac{x}{2} \right) \) can only take values from \([-1, 1]\), it is impossible for it to equal \(-1\) because that would require:
\[ \sin x = -1 \quad \text{and} \quad \sin^2 \left( \frac{x}{2} \right) = 1 \]
This would imply \( \sin^2 \left( \frac{x}{2} \right) = 1\), which occurs when \( \frac{x}{2} = \frac{\pi}{2} + n\pi \) for integers \( n \). Therefore, \( x = \pi + 2n\pi \), which is outside the interval \([-π, π]\).

### Step 4: Conclusion
Since there are no values of \( x \) that satisfy the equation within the given interval, we conclude that:
\[ \text{Number of solutions} = 0 \]

### Final Answer
The number of solutions of the equation \( 1 + \sin x \sin^2 \left( \frac{x}{2} \right) = 0 \) in the interval \([-π, π]\) is \( 0 \).

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