The image you've uploaded shows a problem where you need to determine the correct trigonometric function given certain inputs.

The equation given is:

$f(x) = \sin^{-1}\left(\frac{1}{\sqrt{2}} \right)$

The options seem to ask you to match $x = -\frac{\pi}{4}$, $x = \frac{\pi}{4}$, or $x = -\frac{\pi}{2}$ with $f(x)$.

To solve this:

1. The equation provided is $f(x) = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

   We know that:

   $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$ because $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

2. Therefore, $x = \frac{\pi}{4}$ is the correct answer.

Thus, the correct match is $x = \frac{\pi}{4}$.

Would you like more details or have any questions about this? Here are some follow-up questions related to this topic:

1. How can you determine the values of inverse trigonometric functions for different inputs?
2. What is the domain and range of the $\sin^{-1}(x)$ function?
3. How would you find the $\sin^{-1}(x)$ for values not commonly found on the unit circle?
4. Can you describe the relationship between $\sin(x)$ and $\sin^{-1}(x)$?
5. How would the problem change if the equation was $f(x) = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$?

Tip: Remember that the inverse sine function, $\sin^{-1}(x)$, returns an angle within the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.