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

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `(3 pi)/(2)`
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 need to follow these steps:

### Step 1: Differentiate the function
To determine where the function is increasing, we first need to find its derivative. The function is given by:
\[ f(x) = 3 \sin x - 4 \sin^3 x \]
Differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = 3 \cos x - 12 \sin^2 x \cos x \]
This can be factored as:
\[ f'(x) = 3 \cos x (1 - 4 \sin^2 x) \]

### Step 2: Set the derivative greater than or equal to zero
For the function to be increasing, we need:
\[ f'(x) \geq 0 \]
This leads to:
\[ 3 \cos x (1 - 4 \sin^2 x) \geq 0 \]
Since \( 3 \) is a positive constant, we can simplify this to:
\[ \cos x (1 - 4 \sin^2 x) \geq 0 \]

### Step 3: Analyze the factors
Now we analyze the two factors:
1. \( \cos x \geq 0 \)
2. \( 1 - 4 \sin^2 x \geq 0 \)

#### For \( \cos x \geq 0 \):
This occurs in the intervals:
\[ x \in [2k\pi, (2k+1)\frac{\pi}{2}] \quad \text{for } k \in \mathbb{Z} \]

#### For \( 1 - 4 \sin^2 x \geq 0 \):
Rearranging gives:
\[ 4 \sin^2 x \leq 1 \quad \Rightarrow \quad \sin^2 x \leq \frac{1}{4} \quad \Rightarrow \quad -\frac{1}{2} \leq \sin x \leq \frac{1}{2} \]
The solutions for \( \sin x = \frac{1}{2} \) are:
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \]
The solutions for \( \sin x = -\frac{1}{2} \) are:
\[ x = \frac{7\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2k\pi \]

### Step 4: Find the intervals
The intervals where \( \sin x \) is between \(-\frac{1}{2}\) and \(\frac{1}{2}\) can be determined from the values above. The relevant intervals are:
1. From \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \)
2. From \( \frac{7\pi}{6} \) to \( \frac{11\pi}{6} \)

### Step 5: Determine the length of the intervals
Calculating the lengths of these intervals:
1. Length of \( \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \):
\[ \text{Length} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
2. Length of \( \left[\frac{7\pi}{6}, \frac{11\pi}{6}\right] \):
\[ \text{Length} = \frac{11\pi}{6} - \frac{7\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

### Step 6: Conclusion
The longest interval in which the function is increasing is:
\[ \text{Length} = \frac{2\pi}{3} \]

### Final Answer
The length of the longest interval in which the function \( 3 \sin x - 4 \sin^3 x \) is increasing is \( \frac{2\pi}{3} \).