Question

$\sin \frac{ \pi x}{6} = x^{2} -6x+10$

Answer 1

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 :)