To solve the given problem, we need to find the set of values for the constant \(K\) such that the equation \(\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) = \frac{K\pi}{6}\) has a solution.
- Start by understanding the range of the expression inside the inverse sine function. Since \(\sin^2 x\) can take values from 0 to 1 (because \(-1 \leq \sin x \leq 1\)), the expression \(\frac{1}{1 + \sin^2 x}\) will range from \(\frac{1}{2}\) to 1.
- Therefore, the range of \(\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right)\) will be \([\sin^{-1}(1/2), \sin^{-1}(1)]\), which is \([\pi/6, \pi/2]\).
- The equation \(\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) = \frac{K\pi}{6}\) implies \(\frac{K\pi}{6}\) must lie within this range.
- This implies \(\pi/6 \leq \frac{K\pi}{6} \leq \pi/2\).
- Cancel \(\pi\) from both sides to get \(1 \leq K \leq 3\).
Thus, the set of values for \(K\) for which the equation has a solution is \([1, 3]\). Therefore, the correct answer is: $[1, 3]$.
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