To solve \(\sin^{-1}(\sin(6\pi/7))\), we need to keep in mind that the inverse sine function, \(\sin^{-1}(x)\), gives us an angle in the principal range \([- \frac{\pi}{2}, \frac{\pi}{2}]\), also called the principal value of the inverse sine.

### Step-by-step solution:

1. **Simplify the sine function**:
   \[
   \sin\left(\frac{6\pi}{7}\right)
   \]
   The angle \(\frac{6\pi}{7}\) is between \(\pi/2\) and \(\pi\), which places it in the second quadrant. In the second quadrant, \(\sin(\theta)\) is positive, and for \(\theta = \frac{6\pi}{7}\), the sine is equivalent to the sine of its reference angle:
   \[
   \sin\left(\frac{6\pi}{7}\right) = \sin\left(\pi - \frac{6\pi}{7}\right) = \sin\left(\frac{\pi}{7}\right)
   \]
   
2. **Apply the inverse sine**:
   Now, we apply the inverse sine function, keeping in mind that \(\sin^{-1}(x)\) returns an angle in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\). The sine of \(\frac{6\pi}{7}\) is positive, but \(\frac{6\pi}{7}\) is outside the range of the inverse sine function.

3. **Find the principal value**:
   To bring the angle into the correct range for the inverse sine function, we use the reference angle:
   \[
   \sin^{-1}(\sin(6\pi/7)) = \sin^{-1}(\sin(\pi/7)) = \frac{\pi}{7}
   \]
   This is because \(\pi/7\) lies within the valid range of the inverse sine function.

### Final Answer:
\[
\sin^{-1}(\sin(6\pi/7)) = \frac{\pi}{7}
\]

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Would you like further details or have any questions?

Here are five related questions to explore:

1. What is the principal value of \(\sin^{-1}(\sin(5\pi/4))\)?
2. How do we handle inverse trigonometric functions for angles outside their principal ranges?
3. Can you explain how reference angles work in different quadrants?
4. What is the difference between \(\sin^{-1}(x)\) and \(\frac{1}{\sin(x)}\)?
5. How does the periodicity of sine affect the calculation of inverse functions?

**Tip**: Always check the range of the inverse trigonometric function when solving problems like these, as they restrict the output angle to specific intervals.

Explore how to solve sin^-1(sin(6π/7)) with a detailed step-by-step solution. Learn about inverse trigonometric functions, principal values, and reference angles. This trigonometry problem is ideal for students in Grades 11-12.

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