`f(x)=sin^-1[log_2(x^2/2)]` where [ . ] denotes the greatest integer function.

Explanation

To find the domain of the function \( f(x) = \sin^{-1} \left[ \log_2 \left( \frac{x^2}{2} \right) \right] \), we need to ensure that the argument of the sine inverse function is within its valid range. The sine inverse function, \( \sin^{-1}(y) \), is defined for \( y \) in the interval \([-1, 1]\).

### Step-by-step Solution:

1. **Identify the Argument of the Function**:
We have \( y = \left[ \log_2 \left( \frac{x^2}{2} \right) \right] \). We need to ensure that:
\[ -1 \leq \left[ \log_2 \left( \frac{x^2}{2} \right) \right] \leq 1 \]

2. **Understanding the Greatest Integer Function**:
The greatest integer function \( [y] \) gives the largest integer less than or equal to \( y \). For \( [y] \) to lie between -1 and 1, we have:
\[ [y] = 0 \quad \text{or} \quad [y] = -1 \]

3. **Case 1: \( [y] = 0 \)**:
This means:
\[ 0 \leq \log_2 \left( \frac{x^2}{2} \right) < 1 \]
- From \( 0 \leq \log_2 \left( \frac{x^2}{2} \right) \):
\[ \frac{x^2}{2} \geq 1 \implies x^2 \geq 2 \implies |x| \geq \sqrt{2} \implies x \leq -\sqrt{2} \text{ or } x \geq \sqrt{2} \]
- From \( \log_2 \left( \frac{x^2}{2} \right) < 1 \):
\[ \frac{x^2}{2} < 2 \implies x^2 < 4 \implies |x| < 2 \implies -2 < x < 2 \]

Combining these inequalities, we get:
\[ x \in (-2, -\sqrt{2}] \cup [\sqrt{2}, 2) \]

4. **Case 2: \( [y] = -1 \)**:
This means:
\[ -1 \leq \log_2 \left( \frac{x^2}{2} \right) < 0 \]
- From \( -1 \leq \log_2 \left( \frac{x^2}{2} \right) \):
\[ \frac{x^2}{2} \geq \frac{1}{2} \implies x^2 \geq 1 \implies |x| \geq 1 \implies x \leq -1 \text{ or } x \geq 1 \]
- From \( \log_2 \left( \frac{x^2}{2} \right) < 0 \):
\[ \frac{x^2}{2} < 1 \implies x^2 < 2 \implies |x| < \sqrt{2} \implies -\sqrt{2} < x < \sqrt{2} \]

Combining these inequalities, we get:
\[ x \in (-\sqrt{2}, -1] \cup [1, \sqrt{2}) \]

5. **Final Domain**:
Now, we combine the results from both cases:
- From Case 1: \( (-2, -\sqrt{2}] \cup [\sqrt{2}, 2) \)
- From Case 2: \( (-\sqrt{2}, -1] \cup [1, \sqrt{2}) \)

The intersection of these intervals gives us:
\[ x \in [-2, -1] \cup [1, 2] \]

### Conclusion:
The domain of the function \( f(x) \) is:
\[ \boxed{[-2, -1] \cup [1, 2]} \]