`f(x)=[sinx]` where [.] denotest the greatest integer function is continuous at:

Options

A. `x=(pi)/2`
B. `x=pi`
C. `x=(3pi)/2`
D. `x=2pi`

Answer

To determine where the function \( f(x) = [\sin x] \) (where \([.]\) denotes the greatest integer function) is continuous, we need to analyze the behavior of the sine function and how the greatest integer function affects it.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = [\sin x] \) takes the sine of \( x \) and applies the greatest integer function. The sine function oscillates between -1 and 1.

2. **Identifying Key Points**:
The sine function reaches its maximum value of 1 at \( x = \frac{\pi}{2} \) and its minimum value of -1 at \( x = \frac{3\pi}{2} \). We also note that \( \sin x = 0 \) at \( x = 0, \pi, 2\pi \).

3. **Evaluating Intervals**:
- For \( 0 \leq x < \frac{\pi}{2} \): \( \sin x \) is between 0 and 1, thus \( f(x) = [\sin x] = 0 \).
- At \( x = \frac{\pi}{2} \): \( \sin x = 1 \), hence \( f\left(\frac{\pi}{2}\right) = [1] = 1 \).
- For \( \frac{\pi}{2} < x < \pi \): \( \sin x \) is between 0 and 1, thus \( f(x) = [\sin x] = 0 \).
- At \( x = \pi \): \( \sin x = 0 \), hence \( f(\pi) = [0] = 0 \).
- For \( \pi < x < \frac{3\pi}{2} \): \( \sin x \) is negative, thus \( f(x) = [\sin x] = -1 \).
- At \( x = \frac{3\pi}{2} \): \( \sin x = -1 \), hence \( f\left(\frac{3\pi}{2}\right) = [-1] = -1 \).
- For \( \frac{3\pi}{2} < x < 2\pi \): \( \sin x \) is negative, thus \( f(x) = [\sin x] = -1 \).
- At \( x = 2\pi \): \( \sin x = 0 \), hence \( f(2\pi) = [0] = 0 \).

4. **Checking Continuity**:
- At \( x = 0 \): \( f(0) = 0 \) and the limit from the left is also 0. Thus, continuous.
- At \( x = \frac{\pi}{2} \): \( f\left(\frac{\pi}{2}\right) = 1 \) but the limit from the left is 0. Thus, discontinuous.
- At \( x = \pi \): \( f(\pi) = 0 \) and the limit from the left is 0, and from the right is -1. Thus, discontinuous.
- At \( x = \frac{3\pi}{2} \): \( f\left(\frac{3\pi}{2}\right) = -1 \) and the limit from the left is -1 and from the right is -1. Thus, continuous.
- At \( x = 2\pi \): \( f(2\pi) = 0 \) and the limit from the left is -1. Thus, discontinuous.

5. **Conclusion**:
The function \( f(x) = [\sin x] \) is continuous at \( x = 0 \) and \( x = \frac{3\pi}{2} \).

### Final Answer:
The function \( f(x) = [\sin x] \) is continuous at \( x = 0 \) and \( x = \frac{3\pi}{2} \).