We are tasked with solving the equation:

$\sin^{-1}(x) = \frac{\pi}{6}$

Step 1: Understanding the Equation

The equation states that the arcsine (or inverse sine) of $x$ is equal to $\frac{\pi}{6}$. The arcsine function, $\sin^{-1}(x)$, has a range of $[- \frac{\pi}{2}, \frac{\pi}{2}]$, which means $x$ must be within the domain $[-1, 1]$.

Step 2: Solve for $x$

From the definition of the arcsine function: $\sin^{-1}(x) = \theta \implies \sin(\theta) = x$ In this case: $\sin\left(\frac{\pi}{6}\right) = x$

Step 3: Evaluate $\sin\left(\frac{\pi}{6}\right)$

From the unit circle or trigonometric values: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

Thus: $x = \frac{1}{2}$

Step 4: Determine the Solution Set

The only value for $x$ that satisfies the equation is: $x = \frac{1}{2}$

Final Answer:

The solution is: C. $\{ \frac{1}{2} \}$

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Do you want further explanation or details about arcsine or trigonometric values? Here are some follow-up questions to deepen your understanding:

1. What is the definition of the arcsine function, and how does its range impact solutions?
2. How do we determine the sine of common angles using the unit circle?
3. Can $\sin^{-1}(x)$ ever result in a value outside of $[- \frac{\pi}{2}, \frac{\pi}{2}]$?
4. What is the graphical representation of the arcsine function, and where does $\pi/6$ appear on the graph?
5. How would the solution change if the equation involved $\cos^{-1}(x)$ or $\tan^{-1}(x)$ instead?

Tip: Always remember the ranges and domains of inverse trigonometric functions, as they determine valid solutions.