Number of solutions of equation `sin 9 theta = sin theta` in the interval `[0,2 pi ]` is

Options

A. 16
B. 17
C. 18
D. 15

Answer

To solve the equation \( \sin 9\theta = \sin \theta \) in the interval \( [0, 2\pi] \), we can use the property of the sine function.

### Step-by-step Solution:

1. **Use the sine identity**:
The equation \( \sin A = \sin B \) implies that:
\[ A = B + 2k\pi \quad \text{or} \quad A = \pi - B + 2k\pi \]
where \( k \) is any integer.

For our equation, we set:
\[ 9\theta = \theta + 2k\pi \quad \text{(1)} \]
or
\[ 9\theta = \pi - \theta + 2k\pi \quad \text{(2)} \]

2. **Solve equation (1)**:
Rearranging gives:
\[ 9\theta - \theta = 2k\pi \implies 8\theta = 2k\pi \implies \theta = \frac{k\pi}{4} \]

Now we need to find the values of \( k \) such that \( \theta \) is in the interval \( [0, 2\pi] \):
\[ 0 \leq \frac{k\pi}{4} \leq 2\pi \]
Dividing through by \( \pi \):
\[ 0 \leq \frac{k}{4} \leq 2 \implies 0 \leq k \leq 8 \]
Thus, \( k \) can take values \( 0, 1, 2, \ldots, 8 \) which gives us **9 solutions** from this equation.

3. **Solve equation (2)**:
Rearranging gives:
\[ 9\theta + \theta = \pi + 2k\pi \implies 10\theta = \pi + 2k\pi \implies \theta = \frac{(2k + 1)\pi}{10} \]

Again, we need to find the values of \( k \) such that \( \theta \) is in the interval \( [0, 2\pi] \):
\[ 0 \leq \frac{(2k + 1)\pi}{10} \leq 2\pi \]
Dividing through by \( \pi \):
\[ 0 \leq \frac{(2k + 1)}{10} \leq 2 \implies 0 \leq 2k + 1 \leq 20 \]
This gives:
\[ -1 \leq 2k \leq 19 \implies 0 \leq k \leq 9 \]
Thus, \( k \) can take values \( 0, 1, 2, \ldots, 9 \) which gives us **10 solutions** from this equation.

4. **Total number of solutions**:
Adding the solutions from both equations, we have:
\[ 9 + 10 = 19 \]

Thus, the total number of solutions of the equation \( \sin 9\theta = \sin \theta \) in the interval \( [0, 2\pi] \) is **19**.