Question:

Using the identity $\sin^{-1}(y) + \cos^{-1}(k) = \frac{\pi}{2}$, we get $\sin^{-1}(y) = \sin^{-1}(k)$, so $y = k$. Thus we need $\sin^{-1}\left(x^2-6x+\frac{17}{2}\right) = \sin^{-1}(k)$, which requires both arguments in $[-1,1]$. Let $f(x) = x^2-6x+\frac{17}{2} = (x-3)^2-\frac{1}{2}$. The minimum value is $-\frac{1}{2}$ at $x=3$, and $f(x) \in [-\frac{1}{2}, \infty)$. For real solutions, we need $k \in [-\frac{1}{2}, 1]$ (the intersection of the range of $f$ with $[-1,1]$). When $k = -\frac{1}{2}$, only $x=3$ works (unique solution). For $k \in (-\frac{1}{2}, 1]$, the equation $f(x)=k$ has exactly 2 distinct solutions. The largest value of $k$ giving 2 solutions is $k=1$.