The domain of `f(x)=sin^(-1)[2x^2-3],w h e r e[dot]` denotes the greatest integer function, is `(-sqrt(3/2),sqrt(3/2))` `(-sqrt(3/2),-1)uu(-sqrt(5/2),sqrt(5/2))` `(-sqrt(5/2),sqrt(5/2))` `(-sqrt(5/2),-1)uu(1,sqrt(5/2))`

Options

A. `(-sqrt((3)/(2)),sqrt((3)/(2)))`
B. `(-sqrt((3)/(2)),-1]cup (-sqrt((5)/(2)),sqrt((5)/(2)))`
C. `(-sqrt((5)/(2)),sqrt((5)/(2)))`
D. `(-sqrt((5)/(2)),-1]cup [1,sqrt((5)/(2)))`

Explanation

To find the domain of the function \( f(x) = \sin^{-1}[\lfloor 2x^2 - 3 \rfloor] \), where \( \lfloor \cdot \rfloor \) denotes the greatest integer function, we need to ensure that the argument of the sine inverse function lies within the range of \([-1, 1]\).

### Step-by-Step Solution:

1. **Set up the inequality for the greatest integer function:**
\[ -1 \leq \lfloor 2x^2 - 3 \rfloor \leq 1 \]

2. **Break this down into two inequalities:**
- From \( \lfloor 2x^2 - 3 \rfloor \geq -1 \):
\[ 2x^2 - 3 \geq -1 \implies 2x^2 \geq 2 \implies x^2 \geq 1 \]
- From \( \lfloor 2x^2 - 3 \rfloor \leq 1 \):
\[ 2x^2 - 3 \leq 1 \implies 2x^2 \leq 4 \implies x^2 \leq 2 \]

3. **Combine the results:**
\[ 1 \leq x^2 \leq 2 \]

4. **Take square roots:**
- From \( x^2 \geq 1 \):
\[ |x| \geq 1 \implies x \leq -1 \quad \text{or} \quad x \geq 1 \]
- From \( x^2 \leq 2 \):
\[ |x| \leq \sqrt{2} \implies -\sqrt{2} \leq x \leq \sqrt{2} \]

5. **Combine the intervals:**
- The combined inequalities give us:
\[ x \in (-\sqrt{2}, -1] \cup [1, \sqrt{2}) \]

6. **Final domain:**
- The domain of \( f(x) \) is:
\[ (-\sqrt{2}, -1] \cup [1, \sqrt{2}) \]

### Conclusion:
Thus, the correct option for the domain of \( f(x) \) is:
\[ (-\sqrt{5/2}, -1] \cup [1, \sqrt{5/2}) \]