If `sin (3alpha) /cos(2alpha)` < 0 If `alpha` lies in

A. `(13pi//48,14pi//48)`

B. `(14pi//48,18pi//48)`

C. `(18pi//48,23pi//48)`

D. any of these intervals

Options

A. `(13pi//48,14pi//48)`
B. `(14pi//48,18pi//48)`
C. `(18pi//48,23pi//48)`
D. any of these intervals

Answer

To solve the inequality \(\frac{\sin(3\alpha)}{\cos(2\alpha)} < 0\), we need to analyze the signs of the numerator and denominator separately.

### Step 1: Analyze the numerator \(\sin(3\alpha)\)

Using the triple angle formula for sine, we have:
\[ \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha) \]
We need to find when this expression is negative.

### Step 2: Analyze the denominator \(\cos(2\alpha)\)

Using the double angle formula for cosine, we have:
\[ \cos(2\alpha) = 1 - 2\sin^2(\alpha) \]
We need to find when this expression is positive.

### Step 3: Set up the inequality

The inequality \(\frac{\sin(3\alpha)}{\cos(2\alpha)} < 0\) implies that:
1. \(\sin(3\alpha) < 0\) and \(\cos(2\alpha) > 0\), or
2. \(\sin(3\alpha) > 0\) and \(\cos(2\alpha) < 0\)

### Step 4: Find critical points for \(\sin(3\alpha) < 0\)

The sine function is negative in the intervals:
\[ (3k\pi, (3k+1)\pi) \quad \text{for } k \in \mathbb{Z} \]
Thus, we need to solve:
\[ 3\alpha \in (3k\pi, (3k+1)\pi) \implies \alpha \in \left(k\pi, \frac{(3k+1)\pi}{3}\right) \]

### Step 5: Find critical points for \(\cos(2\alpha) > 0\)

The cosine function is positive in the intervals:
\[ (2k\pi, (2k+1)\pi) \quad \text{for } k \in \mathbb{Z} \]
Thus, we need to solve:
\[ 2\alpha \in (2k\pi, (2k+1)\pi) \implies \alpha \in \left(k\pi, \frac{(2k+1)\pi}{2}\right) \]

### Step 6: Combine intervals

Now, we need to find the intervals where both conditions hold true.

1. For \(\sin(3\alpha) < 0\):
- For \(k=0\): \(\alpha \in (0, \frac{\pi}{3})\)
- For \(k=1\): \(\alpha \in (\pi, \frac{4\pi}{3})\)

2. For \(\cos(2\alpha) > 0\):
- For \(k=0\): \(\alpha \in (0, \frac{\pi}{2})\)
- For \(k=1\): \(\alpha \in (\pi, \frac{3\pi}{2})\)

### Step 7: Find the intersection of intervals

- From \(k=0\):
\[ \alpha \in (0, \frac{\pi}{3}) \cap (0, \frac{\pi}{2}) = (0, \frac{\pi}{3}) \]

- From \(k=1\):
\[ \alpha \in (\pi, \frac{4\pi}{3}) \cap (\pi, \frac{3\pi}{2}) = (\pi, \frac{4\pi}{3}) \]

### Step 8: Convert to the required intervals

We need to convert these intervals into the form given in the options.

1. \(0 < \alpha < \frac{\pi}{3}\) corresponds to \((0, 16\pi/48)\).
2. \(\pi < \alpha < \frac{4\pi}{3}\) corresponds to \((16\pi/48, 24\pi/48)\).

### Conclusion

The intervals where \(\frac{\sin(3\alpha)}{\cos(2\alpha)} < 0\) are:
- \((0, \frac{\pi}{3})\) or \((\pi, \frac{4\pi}{3})\)

Among the options given, the correct answer is:
**B. \((14\pi/48, 18\pi/48)\)**