Identify the correct options :

Options

A. ` (sin 3 alpha )/(cos 2 alpha) gt 0 ` for ` alpha in ((3 pi)/(8), (23 pi)/(48))`
B. ` (sin 3 alpha )/(cos 2 alpha) lt 0 ` for ` alpha in ((13 pi)/(48), (14 pi)/(48))`
C. ` (sin 2 alpha )/(cos alpha) lt 0 ` for ` alpha in (-(pi)/(2), 0)`
D. ` (sin 2 alpha )/(cos alpha) gt 0 ` for ` alpha in ((13 pi)/(48), (14 pi)/(48))`

Answer

To solve the problem, we need to analyze the expression \(\frac{\sin 3\alpha}{\cos 2\alpha}\) and determine the conditions under which it is greater than zero.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ \frac{\sin 3\alpha}{\cos 2\alpha} > 0 \]
This implies that both \(\sin 3\alpha\) and \(\cos 2\alpha\) must either be both positive or both negative.

2. **Finding Conditions for \(\sin 3\alpha > 0\)**:
The sine function is positive in the intervals:
\[ 3\alpha \in (0, \pi) \cup (2\pi, 3\pi) \cup (4\pi, 5\pi) \ldots \]
This translates to:
\[ \alpha \in \left(0, \frac{\pi}{3}\right) \cup \left(\frac{2\pi}{3}, \pi\right) \cup \left(\frac{4\pi}{3}, \frac{5\pi}{3}\right) \ldots \]

3. **Finding Conditions for \(\cos 2\alpha > 0\)**:
The cosine function is positive in the intervals:
\[ 2\alpha \in (-\frac{\pi}{2}, \frac{\pi}{2}) \cup \left(\frac{3\pi}{2}, \frac{5\pi}{2}\right) \ldots \]
This translates to:
\[ \alpha \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \ldots \]

4. **Combining the Conditions**:
We need to find the intersection of the intervals where both conditions hold true:
- From \(\sin 3\alpha > 0\): \(\alpha \in (0, \frac{\pi}{3})\) or \((\frac{2\pi}{3}, \pi)\)
- From \(\cos 2\alpha > 0\): \(\alpha \in (-\frac{\pi}{4}, \frac{\pi}{4})\) or \((\frac{3\pi}{4}, \frac{5\pi}{4})\)

The only overlapping interval is:
\[ \alpha \in (0, \frac{\pi}{4}) \]

5. **Analyzing the Negative Cases**:
Now we consider when both \(\sin 3\alpha < 0\) and \(\cos 2\alpha < 0\):
- \(\sin 3\alpha < 0\) gives us intervals like \((\frac{\pi}{3}, \pi)\) and so on.
- \(\cos 2\alpha < 0\) gives us intervals like \((\frac{\pi}{4}, \frac{3\pi}{4})\).

The overlapping interval here is:
\[ \alpha \in \left(\frac{\pi}{3}, \frac{3\pi}{4}\right) \]

6. **Final Intervals**:
Thus, the complete solution for \(\frac{\sin 3\alpha}{\cos 2\alpha} > 0\) is:
\[ \alpha \in (0, \frac{\pi}{4}) \cup \left(\frac{\pi}{3}, \frac{3\pi}{4}\right) \]

### Conclusion:
Based on the intervals derived, we can now check the provided options to identify which ones fall within these ranges.