The number of solutions of `[sin x+ cos x]=3+ [- sin x]+[-cos x]` (where [.] denotes the greatest integer function), `x in [0, 2pi]`, is

Options

A. 0
B. 4
C. infinite
D. 1

Explanation

To solve the equation \([ \sin x + \cos x ] = 3 + [ - \sin x ] + [ - \cos x ]\) for \(x\) in the interval \([0, 2\pi]\), where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Analyze the left-hand side (LHS)
The LHS is \([ \sin x + \cos x ]\). The maximum value of \(\sin x + \cos x\) occurs at \(x = \frac{\pi}{4}\), where:
\[ \sin x + \cos x = \sqrt{2} \]
The minimum value occurs at \(x = \frac{5\pi}{4}\), where:
\[ \sin x + \cos x = -\sqrt{2} \]
Thus, the range of \(\sin x + \cos x\) is \([- \sqrt{2}, \sqrt{2}]\). Therefore, the maximum value of the LHS is:
\[ [\sin x + \cos x] \leq 1 \]

### Step 2: Analyze the right-hand side (RHS)
The RHS is \(3 + [ - \sin x ] + [ - \cos x ]\). The values of \([- \sin x]\) and \([- \cos x]\) can be determined as follows:
- For \(x \in [0, \frac{\pi}{2}]\), \(\sin x\) and \(\cos x\) are both in \([0, 1]\), hence \([- \sin x] = -1\) and \([- \cos x] = -1\).
- For \(x \in [\frac{\pi}{2}, \frac{3\pi}{2}]\), \(\sin x\) is in \([-1, 0]\) and \(\cos x\) is in \([-1, 0]\), hence \([- \sin x] = 0\) and \([- \cos x] = 0\).
- For \(x \in [\frac{3\pi}{2}, 2\pi]\), \(\sin x\) and \(\cos x\) are both in \([-1, 0]\), hence \([- \sin x] = 0\) and \([- \cos x] = 0\).

Thus, the RHS can be computed:
- For \(x \in [0, \frac{\pi}{2}]\):
\[ RHS = 3 + (-1) + (-1) = 1 \]
- For \(x \in [\frac{\pi}{2}, 2\pi]\):
\[ RHS = 3 + 0 + 0 = 3 \]

### Step 3: Set up the equation
Now we equate LHS and RHS:
1. For \(x \in [0, \frac{\pi}{2}]\):
\[ [\sin x + \cos x] = 1 \]
This implies:
\[ \sin x + \cos x = 1 \]
The only solution in this interval is \(x = \frac{\pi}{4}\).

2. For \(x \in [\frac{\pi}{2}, 2\pi]\):
\[ [\sin x + \cos x] = 3 \]
This is impossible since the maximum value of LHS is 1.

### Step 4: Count the solutions
The only solution we found is \(x = \frac{\pi}{4}\) in the interval \([0, \frac{\pi}{2}]\). Therefore, the total number of solutions in the interval \([0, 2\pi]\) is:

\[ \text{Number of solutions} = 1 \]

### Final Answer
The number of solutions of the equation is \(1\).