The true set of values of 'K' for which $\sin^{-1}\left(\dfrac{1}{1 + \sin^2 x}\right) = \dfrac{K\pi}{6}$ may have a solution is

We know that $0 \le \sin^2 x \le 1$ for all real $x$.

Adding $1$ to all parts, we get $1 \le 1 + \sin^2 x \le 2$.

Taking the reciprocal, we have $\dfrac{1}{2} \le \dfrac{1}{1 + \sin^2 x} \le 1$.

Since $\sin^{-1} y$ is an increasing function for $y \in [-1, 1]$, we can apply it to the inequality:

$\sin^{-1}\left(\dfrac{1}{2}\right) \le \sin^{-1}\left(\dfrac{1}{1 + \sin^2 x}\right) \le \sin^{-1}(1)$

Substituting the known values of the inverse sine function:

$\dfrac{\pi}{6} \le \sin^{-1}\left(\dfrac{1}{1 + \sin^2 x}\right) \le \dfrac{\pi}{2}$

Given the equation $\sin^{-1}\left(\dfrac{1}{1 + \sin^2 x}\right) = \dfrac{K\pi}{6}$, for a solution to exist, the right-hand side must lie in the range of the left-hand side:

$\dfrac{\pi}{6} \le \dfrac{K\pi}{6} \le \dfrac{\pi}{2}$

Dividing the entire inequality by $\dfrac{\pi}{6}$, we obtain:

$1 \le K \le 3$

Thus, the true set of values of $K$ is $[1, 3]$.

Answer: $[1, 3]$