The number of solutions of the equation `sin x . Sin 2x. Sin 3x=1` in `[0,2pi]` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \( \sin x \cdot \sin 2x \cdot \sin 3x = 1 \) in the interval \( [0, 2\pi] \), we need to analyze the conditions under which the product of these sine functions can equal 1.

### Step-by-Step Solution:

1. **Understanding the Range of Sine Functions**:
- The sine function, \( \sin x \), can take values from -1 to 1. Therefore, the product \( \sin x \cdot \sin 2x \cdot \sin 3x \) can also take values in the range of -1 to 1.
- For the product to equal 1, each sine function must individually equal 1, since the maximum value of a product of numbers occurs when all numbers are at their maximum.

2. **Finding Values for Each Sine Function**:
- The sine function equals 1 at specific points:
- \( \sin x = 1 \) when \( x = \frac{\pi}{2} + 2k\pi \) for any integer \( k \).
- \( \sin 2x = 1 \) when \( 2x = \frac{\pi}{2} + 2k\pi \) which simplifies to \( x = \frac{\pi}{4} + k\pi \).
- \( \sin 3x = 1 \) when \( 3x = \frac{\pi}{2} + 2k\pi \) which simplifies to \( x = \frac{\pi}{6} + \frac{2k\pi}{3} \).

3. **Checking for Simultaneous Solutions**:
- We need to find values of \( x \) such that all three sine functions equal 1 at the same time.
- The first sine function \( \sin x = 1 \) gives \( x = \frac{\pi}{2} \).
- For \( \sin 2x = 1 \), substituting \( x = \frac{\pi}{2} \) gives \( 2x = \pi \), which is valid since \( \sin \pi = 0 \).
- For \( \sin 3x = 1 \), substituting \( x = \frac{\pi}{2} \) gives \( 3x = \frac{3\pi}{2} \), which is also valid since \( \sin \frac{3\pi}{2} = -1 \).

4. **Conclusion**:
- Since we cannot find a single value of \( x \) in the interval \( [0, 2\pi] \) where all three sine functions equal 1 simultaneously, we conclude that there are no solutions to the equation \( \sin x \cdot \sin 2x \cdot \sin 3x = 1 \) in the given interval.

Thus, the number of solutions is **0**.