To solve the inequality \(\frac{\sin(3\theta)}{\cos(2\theta)} < 0\), we need to analyze when the numerator and denominator have opposite signs. This can happen in two scenarios:
### Step 1: Identify the cases
1. **Case 1:** \(\sin(3\theta) < 0\) and \(\cos(2\theta) > 0\)
2. **Case 2:** \(\sin(3\theta) > 0\) and \(\cos(2\theta) < 0\)
### Step 2: Analyze Case 1
**For Case 1:**
- \(\sin(3\theta) < 0\) occurs in the intervals where \(3\theta\) is in the third and fourth quadrants:
\[
3\theta \in (\pi, 2\pi) \Rightarrow \theta \in \left(\frac{\pi}{3}, \frac{2\pi}{3}\right)
\]
- \(\cos(2\theta) > 0\) occurs in the intervals where \(2\theta\) is in the first and fourth quadrants:
\[
2\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow \theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)
\]
### Step 3: Combine results from Case 1
From Case 1, we have:
- \(\theta \in \left(\frac{\pi}{3}, \frac{2\pi}{3}\right)\)
- \(\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)
The intersection of these intervals is empty, so we move to Case 2.
### Step 4: Analyze Case 2
**For Case 2:**
- \(\sin(3\theta) > 0\) occurs in the intervals where \(3\theta\) is in the first and second quadrants:
\[
3\theta \in (0, \pi) \Rightarrow \theta \in \left(0, \frac{\pi}{3}\right)
\]
- \(\cos(2\theta) < 0\) occurs in the intervals where \(2\theta\) is in the second and third quadrants:
\[
2\theta \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \Rightarrow \theta \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)
\]
### Step 5: Combine results from Case 2
From Case 2, we have:
- \(\theta \in \left(0, \frac{\pi}{3}\right)\)
- \(\theta \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)\)
The intersection of these intervals is:
\[
\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)
\]
### Final Result
Thus, the final range for \(\theta\) where \(\frac{\sin(3\theta)}{\cos(2\theta)} < 0\) is:
\[
\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)
\]
