If `m, n in N`( `n> m`), then number of solutions of the equation `n|sinx|=m|sinx| ` in `[0, 2pi]` is

Options

A. m
B. n
C. mn
D. 3

Answer

To solve the equation \( n|\sin x| = m|\sin x| \) for \( x \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ n|\sin x| = m|\sin x| \]
Rearranging gives us:
\[ n|\sin x| - m|\sin x| = 0 \]
Factoring out \(|\sin x|\):
\[ (n - m)|\sin x| = 0 \]

### Step 2: Analyzing the Factors
The equation \((n - m)|\sin x| = 0\) implies that either:
1. \(n - m = 0\) or
2. \(|\sin x| = 0\)

Given that \(n > m\), we know that \(n - m \neq 0\). Therefore, we must have:
\[ |\sin x| = 0 \]

### Step 3: Finding Solutions for \(|\sin x| = 0\)
The equation \(|\sin x| = 0\) implies:
\[ \sin x = 0 \]
The general solutions for \(\sin x = 0\) are:
\[ x = n\pi \quad \text{where } n \text{ is an integer} \]

### Step 4: Identifying Valid Solutions in the Interval
We need to find the values of \(x\) within the interval \([0, 2\pi]\):
- For \(n = 0\): \(x = 0\)
- For \(n = 1\): \(x = \pi\)
- For \(n = 2\): \(x = 2\pi\)

Thus, the solutions in the interval \([0, 2\pi]\) are:
\[ x = 0, \pi, 2\pi \]

### Step 5: Counting the Solutions
We have found three solutions:
1. \(x = 0\)
2. \(x = \pi\)
3. \(x = 2\pi\)

Thus, the total number of solutions to the equation \(n|\sin x| = m|\sin x|\) in the interval \([0, 2\pi]\) is:
\[ \boxed{3} \]