The value of `[sin x]+[1+sin x]+[2+sin x]` in `x in (pi,3pi//2]` can be ([.] is the greatest integer function) can be.

Options

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

Explanation

To solve the problem, we need to evaluate the expression \([ \sin x ] + [ 1 + \sin x ] + [ 2 + \sin x ]\) for \(x \in (\pi, \frac{3\pi}{2}]\), where \([ . ]\) denotes the greatest integer function (also known as the floor function).

### Step 1: Determine the range of \(\sin x\) for \(x \in (\pi, \frac{3\pi}{2}]\)

In the interval \(x \in (\pi, \frac{3\pi}{2}]\), the sine function is negative. Specifically, we know:
- At \(x = \pi\), \(\sin x = 0\).
- At \(x = \frac{3\pi}{2}\), \(\sin x = -1\).

Thus, for \(x\) in the interval \((\pi, \frac{3\pi}{2}]\), the value of \(\sin x\) varies from \(0\) to \(-1\). Therefore, we can conclude:
\[ -1 < \sin x < 0 \]

### Step 2: Evaluate \([ \sin x ]\)

Since \(\sin x\) is negative and lies between \(-1\) and \(0\), we have:
\[ [ \sin x ] = -1 \]

### Step 3: Evaluate \([ 1 + \sin x ]\)

Next, we evaluate \(1 + \sin x\):
\[ 1 + \sin x \text{ varies from } 1 + 0 = 1 \text{ to } 1 - 1 = 0 \]
Thus, \(0 < 1 + \sin x < 1\), which gives us:
\[ [ 1 + \sin x ] = 0 \]

### Step 4: Evaluate \([ 2 + \sin x ]\)

Now, we evaluate \(2 + \sin x\):
\[ 2 + \sin x \text{ varies from } 2 + 0 = 2 \text{ to } 2 - 1 = 1 \]
Thus, \(1 < 2 + \sin x < 2\), which gives us:
\[ [ 2 + \sin x ] = 1 \]

### Step 5: Combine the results

Now we can combine all the evaluated parts:
\[ [ \sin x ] + [ 1 + \sin x ] + [ 2 + \sin x ] = -1 + 0 + 1 = 0 \]

### Conclusion

The value of \([ \sin x ] + [ 1 + \sin x ] + [ 2 + \sin x ]\) for \(x \in (\pi, \frac{3\pi}{2}]\) is \(0\).

### Final Answer

Thus, the answer is \(0\).

---