To determine the domain of inverse trigonometric functions, ensure that the input expression lies within the domain of the original function. For $ \sin^{-1}(y) $, the valid input range is $-1 \leq y \leq 1$.
We are given the expression $ \sin^{-1}(2x^2 - 3) $, and we need to find the domain for which this function is defined. The domain of the inverse sine function is limited to the range $[-1, 1]$. Thus, we need the expression inside the inverse sine function, $2x^2 - 3$, to lie within this range.
\[
-1 \leq 2x^2 - 3 \leq 1
\]
First, solve for $x^2$:
\[
-1 \leq 2x^2 - 3 \leq 1
\]
\[
2 \leq 2x^2 \leq 4
\]
\[
1 \leq x^2 \leq 2
\]
Taking square roots on both sides:
\[
\sqrt{1} \leq |x| \leq \sqrt{2}
\]
\[
1 \leq |x| \leq \sqrt{2}
\]
Thus, the values of $x$ lie between $[-\sqrt{2}, -1]$ and $[1, \sqrt{2}]$. The domain of the function is:
\[
[-\sqrt{2}, -1] \cup [1, \sqrt{2}]
\]
