The range of the sine function is \([-1,1]\), so:
\(x^2-6x+10\geq-1 \wedge x^2-6x+10 \leq1\\\\
x^2-6x+10\geq-1\\
x^2-6x+11\geq0\\
\Delta=(-6)^2-4\cdot1\cdot11=36-44=-8\\
x\in \mathbb{R}\\\\
x^2-6x+10\leq1\\
x^2-6x+9\leq0\\
(x-3)^2\leq0\\
(x-3)^2=0\\
x-3=0\\
x=3\\\\
x\in \mathbb{R} \wedge x=3\\
x=3\)
So 3 is the only possible value the function \(x^2-6x+10\) can take as an argument. Let's see if 3 is a solution.
\(\sin\dfrac{\pi\cdot3}{6}=3^2-6\cdot3+10\\
\sin \dfrac{\pi}{2}=9-18+10\\
1=1\)
Therefore it is :)
