\sin ( \pi x)/(6) = x^(2) -6x+10

Question

`\sin ( \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(\pi\cdot3)/(6)=3^2-6\cdot3+10\\ \sin (\pi)/(2)=9-18+10\\ 1=1`

Therefore it is :)