Show that the equation , ` sin x =[1+ sin x]+[1- cos x]` has no solution for ` x in R `. ( where [.] represents greatest integers functions ).

Answer

To show that the equation \( \sin x = [1 + \sin x] + [1 - \cos x] \) has no solution for \( x \in \mathbb{R} \), we will analyze the equation step by step.

### Step 1: Understand the components of the equation
The equation involves the sine and cosine functions, which are periodic with a period of \( 2\pi \). The greatest integer function (denoted by square brackets) returns the largest integer less than or equal to the argument.

### Step 2: Rewrite the equation
We can rewrite the equation as follows:
\[ \sin x = [1 + \sin x] + [1 - \cos x] \]
This can be expressed as:
\[ \sin x = [1 + \sin x] + [1 - \cos x] \]

### Step 3: Analyze the ranges of \( \sin x \) and \( \cos x \)
The values of \( \sin x \) range from -1 to 1, and \( \cos x \) also ranges from -1 to 1. Thus:
- \( 1 + \sin x \) ranges from 0 to 2.
- \( 1 - \cos x \) ranges from 0 to 2.

### Step 4: Determine the values of the greatest integer functions
Now, we need to evaluate the greatest integer functions:
- Since \( 1 + \sin x \) ranges from 0 to 2, \( [1 + \sin x] \) can be either 0 or 1.
- Since \( 1 - \cos x \) also ranges from 0 to 2, \( [1 - \cos x] \) can also be either 0 or 1.

### Step 5: Evaluate possible cases
We can evaluate the possible cases for \( [1 + \sin x] \) and \( [1 - \cos x] \):
1. **Case 1**: If \( [1 + \sin x] = 0 \) and \( [1 - \cos x] = 0 \):
\[ \sin x = 0 + 0 = 0 \]
This is true for \( x = n\pi \) where \( n \) is an integer.

2. **Case 2**: If \( [1 + \sin x] = 0 \) and \( [1 - \cos x] = 1 \):
\[ \sin x = 0 + 1 = 1 \]
This is true for \( x = \frac{\pi}{2} + 2n\pi \).

3. **Case 3**: If \( [1 + \sin x] = 1 \) and \( [1 - \cos x] = 0 \):
\[ \sin x = 1 + 0 = 1 \]
This is true for \( x = \frac{\pi}{2} + 2n\pi \).

4. **Case 4**: If \( [1 + \sin x] = 1 \) and \( [1 - \cos x] = 1 \):
\[ \sin x = 1 + 1 = 2 \]
This is impossible since \( \sin x \) cannot exceed 1.

### Step 6: Conclusion
From the analysis of all cases, we find that:
- In cases 1, 2, and 3, we have conditions that do not hold simultaneously for any \( x \).
- The only possible value for \( \sin x \) that satisfies the equation is 2, which is not possible.

Thus, we conclude that the equation \( \sin x = [1 + \sin x] + [1 - \cos x] \) has no solution for \( x \in \mathbb{R} \).