The length of the longest interval, in which the function `3sin x-4 sin^3x` is increasing is

Options

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

Explanation

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

### Step 1: Differentiate the function
We start by finding the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(3\sin x - 4\sin^3 x) \]
Using the chain rule and the derivative of sine:
\[ f'(x) = 3\cos x - 12\sin^2 x \cos x \]
Factoring out \( \cos x \):
\[ f'(x) = \cos x (3 - 12\sin^2 x) \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we set the derivative greater than zero:
\[ \cos x (3 - 12\sin^2 x) > 0 \]
This inequality holds when both factors are positive or both are negative.

### Step 3: Analyze the factors
1. **For \( \cos x > 0 \)**:
- This occurs in the intervals \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, \frac{5\pi}{2}) \) within one period \( [0, 2\pi] \).

2. **For \( 3 - 12\sin^2 x > 0 \)**:
- Rearranging gives \( \sin^2 x < \frac{1}{4} \).
- Thus, \( -\frac{1}{2} < \sin x < \frac{1}{2} \).
- The corresponding intervals for \( x \) are:
- \( x \in (-\frac{\pi}{6}, \frac{\pi}{6}) \) and \( x \in (\frac{5\pi}{6}, \frac{7\pi}{6}) \).

### Step 4: Find the intersection of intervals
Now we find where both conditions hold:
- From \( \cos x > 0 \): \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)
- From \( 3 - 12\sin^2 x > 0 \): \( (-\frac{\pi}{6}, \frac{\pi}{6}) \)

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

### Step 5: Calculate the length of the interval
The length of the interval \( (-\frac{\pi}{6}, \frac{\pi}{6}) \) is:
\[ \text{Length} = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Final Answer
Thus, the length of the longest interval in which the function \( 3\sin x - 4\sin^3 x \) is increasing is:
\[ \frac{\pi}{3} \]

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