h(x) is increasing for `x in `

Options

A. `(-(pi)/(4),(pi)/(4))`
B. `(-(pi)/(2),-(pi)/(4))uu(pi)/(4),(pi)/(2))`
C. `(-(pi)/(4),0)uu(pi)/(4),(pi)/(2))`
D. `(-(pi)/(2),-(pi)/(4))uu(0,(pi)/(4))`

Explanation

To determine the intervals where the function \( h(x) \) is increasing, we need to analyze the derivative \( h'(x) \). A function is increasing on an interval if its derivative is greater than zero on that interval.

### Step 1: Find the derivative
Given that \( h(x) = \sin(2x) \), we first compute the derivative:
\[ h'(x) = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \]

### Step 2: Set the derivative greater than zero
To find where \( h(x) \) is increasing, we set the derivative greater than zero:
\[ 2\cos(2x) > 0 \]
This simplifies to:
\[ \cos(2x) > 0 \]

### Step 3: Determine the intervals for \( \cos(2x) > 0 \)
The cosine function is positive in the intervals:
\[ 2x \in (2n\pi, (2n+1)\pi) \quad \text{for } n \in \mathbb{Z} \]
Dividing by 2 gives:
\[ x \in \left(n\pi, \frac{(2n+1)\pi}{2}\right) \quad \text{for } n \in \mathbb{Z} \]

### Step 4: Identify specific intervals
For \( n = 0 \):
\[ x \in (0, \frac{\pi}{2}) \]
For \( n = -1 \):
\[ x \in \left(-\frac{\pi}{2}, 0\right) \]

### Step 5: Combine the intervals
Combining the intervals from both cases, we find:
\[ x \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right) \]

### Final Answer
Thus, the function \( h(x) \) is increasing for:
\[ x \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right) \]