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

Options

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

Answer

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

### Step 1: Find the derivative of the function
To determine where the function is increasing, we first find its derivative:
\[ A' = \frac{d}{dx}(3\sin x - 4\sin^3 x) \]
Using the chain rule and the power rule, we differentiate:
\[ A' = 3\cos x - 12\sin^2 x \cos x \]
This can be factored as:
\[ A' = 3\cos x(1 - 4\sin^2 x) \]

### Step 2: Set the derivative greater than zero
For the function to be increasing, we need:
\[ A' > 0 \implies 3\cos x(1 - 4\sin^2 x) > 0 \]
This implies two conditions must hold:
1. \( \cos x > 0 \)
2. \( 1 - 4\sin^2 x > 0 \)

### Step 3: Solve the inequalities
**Condition 1: \( \cos x > 0 \)**
This occurs in the intervals:
\[ -\frac{\pi}{2} < x < \frac{\pi}{2} \]

**Condition 2: \( 1 - 4\sin^2 x > 0 \)**
Rearranging gives:
\[ 4\sin^2 x < 1 \implies \sin^2 x < \frac{1}{4} \implies -\frac{1}{2} < \sin x < \frac{1}{2} \]
The values of \( x \) satisfying this condition can be found by considering the sine function:
\[ \sin x = \frac{1}{2} \implies x = \frac{\pi}{6} \quad \text{and} \quad x = -\frac{\pi}{6} \]

### Step 4: Combine the intervals
Now we combine the intervals from both conditions:
- From \( \cos x > 0 \): \( -\frac{\pi}{2} < x < \frac{\pi}{2} \)
- From \( -\frac{1}{2} < \sin x < \frac{1}{2} \): \( -\frac{\pi}{6} < x < \frac{\pi}{6} \)

The intersection of these intervals gives:
\[ -\frac{\pi}{6} < x < \frac{\pi}{6} \]

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

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

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