The given mathematical expression is $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$, which represents the arcsine or inverse sine function.

To solve this:

1. Recall that the sine of an angle gives the ratio of the opposite side to the hypotenuse in a right triangle. The arcsine function $\sin^{-1}(x)$ gives the angle $\theta$ whose sine is $x$.

2. From the unit circle: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.$

3. Therefore: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.$

The solution is: $\boxed{\frac{\pi}{3}}$

Would you like further details on the unit circle or inverse trigonometric functions?

Related Questions:

1. What is the domain and range of the arcsine function?
2. How is $\sin^{-1}(x)$ different from $\csc^{-1}(x)$?
3. Can you explain why $\sin^{-1}(x)$ only has one principal value?
4. What are some real-world applications of inverse trigonometric functions?
5. How do you solve $\sin^{-1}(-x)$?

Tip:

When working with inverse trigonometric functions, always consider their principal branches, which determine the range of possible solutions. For $\sin^{-1}(x)$, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.