The function `f(x) =4sin^(3)x-6sin^(2)x +12 sinx + 100` is strictly

Options

A. increasing in `pi, (3pi)/2`
B. decreasing in `(pi/2,pi)`
C. decreasing in `[-pi/2, pi/2]`
D. decreasing in `[0,pi/2]`

Explanation

To determine the intervals where the function \( f(x) = 4\sin^3 x - 6\sin^2 x + 12\sin x + 100 \) is strictly increasing or decreasing, we need to follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(4\sin^3 x - 6\sin^2 x + 12\sin x + 100) \]

Using the chain rule and the power rule, we differentiate each term:

\[ f'(x) = 4 \cdot 3\sin^2 x \cdot \cos x - 6 \cdot 2\sin x \cdot \cos x + 12 \cdot \cos x \]

This simplifies to:

\[ f'(x) = 12\sin^2 x \cos x - 12\sin x \cos x + 12\cos x \]

Factoring out \( 12\cos x \):

\[ f'(x) = 12\cos x (\sin^2 x - \sin x + 1) \]

### Step 2: Analyze the derivative
Next, we need to analyze the sign of \( f'(x) \).

1. **Identify the factors**:
- \( 12\cos x \) is positive when \( \cos x > 0 \) (i.e., in the intervals \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)).
- The quadratic \( \sin^2 x - \sin x + 1 \) can be analyzed for its roots. The discriminant \( D \) of this quadratic is:

\[ D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]

Since the discriminant is negative, the quadratic has no real roots and is always positive.

### Step 3: Determine intervals of increase and decrease
Given that \( \sin^2 x - \sin x + 1 > 0 \) for all \( x \), the sign of \( f'(x) \) depends solely on \( \cos x \):

- **Increasing**: \( f'(x) > 0 \) when \( \cos x > 0 \) (i.e., \( x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \)).
- **Decreasing**: \( f'(x) < 0 \) when \( \cos x < 0 \) (i.e., \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \)).

### Conclusion
Thus, the function \( f(x) \) is strictly increasing in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) and strictly decreasing in the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).