The number of real solutions of the equation x-sinx=0, is

Options

A. 0
B. 1
C. 2
D. infinitely many

Explanation

To determine the number of real solutions of the equation \( x - \sin x = 0 \), we can follow these steps:

### Step 1: Define the Function
Let \( f(x) = x - \sin x \).

### Step 2: Differentiate the Function
We need to find the derivative of \( f(x) \):
\[ f'(x) = 1 - \cos x \]

### Step 3: Analyze the Derivative
The function \( \cos x \) oscillates between -1 and 1, which means \( 1 - \cos x \) is always non-negative:
\[ f'(x) \geq 0 \quad \text{for all } x \]
This implies that \( f(x) \) is a non-decreasing function.

### Step 4: Identify Critical Points
The derivative \( f'(x) = 0 \) when \( \cos x = 1 \). This occurs at:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
At these points, \( f(x) \) could change its behavior.

### Step 5: Evaluate the Function at Critical Points
Now, we will evaluate \( f(x) \) at \( x = 0 \):
\[ f(0) = 0 - \sin(0) = 0 \]
Thus, \( f(0) = 0 \).

### Step 6: Analyze Behavior of the Function
- For \( x > 0 \):
Since \( f'(x) \geq 0 \), \( f(x) \) is increasing. Therefore, for \( x > 0 \), \( f(x) > 0 \).

- For \( x < 0 \):
Similarly, since \( f(x) \) is increasing and \( f(0) = 0 \), it follows that for \( x < 0 \), \( f(x) < 0 \).

### Step 7: Conclusion
Since \( f(x) \) is strictly increasing and crosses the x-axis only at \( x = 0 \), we conclude that the equation \( x - \sin x = 0 \) has exactly one real solution:
\[ \text{The number of real solutions is } 1. \]