What is die length of the longest interval in which the function `f(x) = 3 sinx - 4 sin^3 x` is increasing?

Options

A. A. `pi//3`
B. B. `pi//2`
C. C. `3pi//2`
D. D. `pi`

Answer

To find the length of the longest interval in which the function \( f(x) = 3 \sin x - 4 \sin^3 x \) is increasing, we can follow these steps:

### Step 1: Identify the function and its derivative
The function given is:
\[ f(x) = 3 \sin x - 4 \sin^3 x \]
To determine where the function is increasing, we need to find its derivative \( f'(x) \).

### Step 2: Differentiate the function
Using the chain rule and the power rule, we differentiate \( f(x) \):
\[ f'(x) = 3 \cos x - 12 \sin^2 x \cos x \]
This can be factored as:
\[ f'(x) = 3 \cos x (1 - 4 \sin^2 x) \]

### Step 3: Set the derivative greater than zero
To find where the function is increasing, we need to solve the inequality:
\[ f'(x) > 0 \]
This gives us:
\[ 3 \cos x (1 - 4 \sin^2 x) > 0 \]
This means both factors must be positive or both must be negative.

### Step 4: Analyze the factors
1. **For \( 3 \cos x > 0 \)**:
- This occurs when \( \cos x > 0 \), which is in the intervals:
\[ x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \quad \text{and} \quad x \in (2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}) \text{ for } n \in \mathbb{Z} \]

2. **For \( 1 - 4 \sin^2 x > 0 \)**:
- This simplifies to:
\[ \sin^2 x < \frac{1}{4} \implies -\frac{1}{2} < \sin x < \frac{1}{2} \]
- The intervals for \( \sin x \) being between \(-\frac{1}{2}\) and \(\frac{1}{2}\) are:
\[ x \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \quad \text{and} \quad x \in \left(\frac{5\pi}{6}, \frac{7\pi}{6}\right) \]

### Step 5: Find the intersection of intervals
Now we need to find the intersection of the intervals where both conditions hold:
- From \( 3 \cos x > 0 \):
- \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)

- From \( 1 - 4 \sin^2 x > 0 \):
- \( (-\frac{\pi}{6}, \frac{\pi}{6}) \)

The intersection of these intervals is:
\[ (-\frac{\pi}{6}, \frac{\pi}{6}) \]

### Step 6: Calculate the length of the interval
The length of the interval \( (-\frac{\pi}{6}, \frac{\pi}{6}) \) is:
\[ \text{Length} = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Conclusion
Thus, the length of the longest interval in which the function \( f(x) \) is increasing is:
\[ \boxed{\frac{\pi}{3}} \]

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