Question

Find a function ω(x) such that the function f(x) = sin(ω(x)) has the roots at π, π2

Answer 1

So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.

$f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}$