The number of roots of the equation `sin(2x+pi/18) cos(2x-pi/9)=-1/4` in `[0, 2pi]` is

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the equation \( \sin(2x + \frac{\pi}{18}) \cos(2x - \frac{\pi}{9}) = -\frac{1}{4} \) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ \sin(2x + \frac{\pi}{18}) \cos(2x - \frac{\pi}{9}) = -\frac{1}{4} \]

### Step 2: Use the product-to-sum identities
We can use the product-to-sum identities to rewrite the left-hand side:
\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
Let \( A = 2x + \frac{\pi}{18} \) and \( B = 2x - \frac{\pi}{9} \).

Calculating \( A + B \) and \( A - B \):
\[ A + B = (2x + \frac{\pi}{18}) + (2x - \frac{\pi}{9}) = 4x + \frac{\pi}{18} - \frac{2\pi}{18} = 4x - \frac{\pi}{18} \]
\[ A - B = (2x + \frac{\pi}{18}) - (2x - \frac{\pi}{9}) = \frac{\pi}{18} + \frac{2\pi}{18} = \frac{3\pi}{18} = \frac{\pi}{6} \]

Thus, we can rewrite the equation as:
\[ \frac{1}{2} \left[ \sin(4x - \frac{\pi}{18}) + \sin(\frac{\pi}{6}) \right] = -\frac{1}{4} \]

### Step 3: Simplify the equation
Since \( \sin(\frac{\pi}{6}) = \frac{1}{2} \), we have:
\[ \frac{1}{2} \left[ \sin(4x - \frac{\pi}{18}) + \frac{1}{2} \right] = -\frac{1}{4} \]
Multiplying both sides by 2:
\[ \sin(4x - \frac{\pi}{18}) + \frac{1}{2} = -\frac{1}{2} \]
Subtracting \(\frac{1}{2}\) from both sides:
\[ \sin(4x - \frac{\pi}{18}) = -1 \]

### Step 4: Solve for \(4x - \frac{\pi}{18}\)
The sine function equals -1 at:
\[ 4x - \frac{\pi}{18} = \frac{3\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

### Step 5: Solve for \(x\)
Rearranging gives:
\[ 4x = \frac{3\pi}{2} + \frac{\pi}{18} + 2k\pi \]
Finding a common denominator (which is 18):
\[ 4x = \frac{27\pi}{18} + \frac{\pi}{18} + 2k\pi = \frac{28\pi}{18} + 2k\pi = \frac{14\pi}{9} + 2k\pi \]
Dividing by 4:
\[ x = \frac{14\pi}{36} + \frac{k\pi}{2} = \frac{7\pi}{18} + \frac{k\pi}{2} \]

### Step 6: Find values of \(x\) in \([0, 2\pi]\)
Now we need to find integer values of \(k\) such that \(x\) lies in \([0, 2\pi]\):
1. For \(k = 0\):
\[ x = \frac{7\pi}{18} \quad \text{(valid)} \]
2. For \(k = 1\):
\[ x = \frac{7\pi}{18} + \frac{\pi}{2} = \frac{7\pi}{18} + \frac{9\pi}{18} = \frac{16\pi}{18} = \frac{8\pi}{9} \quad \text{(valid)} \]
3. For \(k = 2\):
\[ x = \frac{7\pi}{18} + \pi = \frac{7\pi}{18} + \frac{18\pi}{18} = \frac{25\pi}{18} \quad \text{(valid)} \]
4. For \(k = 3\):
\[ x = \frac{7\pi}{18} + \frac{3\pi}{2} = \frac{7\pi}{18} + \frac{27\pi}{18} = \frac{34\pi}{18} = \frac{17\pi}{9} \quad \text{(not valid as it exceeds } 2\pi\text{)} \]

Thus, we have found three valid solutions for \(x\) in the interval \([0, 2\pi]\):
- \(x = \frac{7\pi}{18}\)
- \(x = \frac{8\pi}{9}\)
- \(x = \frac{25\pi}{18}\)

### Conclusion
The number of roots of the equation \( \sin(2x + \frac{\pi}{18}) \cos(2x - \frac{\pi}{9}) = -\frac{1}{4} \) in the interval \([0, 2\pi]\) is **3**.