[Math] How to prove that $1-\sin(x)$ is one-to-one or not

Question

I'm currently at the very beginning of my calculus course and I've run into an immediate issue in that I'm not entirely sure how to verify when a function is one-to-one. I know that they're considered one-to-one if $f(x)=f(y)$ when $y=x$, but when I apply this to $f(x)=1-\sin(x)$, the function appears as one-to-one even though I know it isn't unless the domain is restricted.

My work thus far is as follows:

Verify if $f(x)=1-\sin(x)$ is one-to-one

$f(a)=1-\sin(a)$ ;
$f(b)=1-\sin(b)$

$\Rightarrow1-\sin(a)=1-\sin(b)$

$\Rightarrow-\sin(a)=-\sin(b) \rightarrow \sin(a)=\sin(b)$

$\Rightarrow \arcsin(\sin(a))=\arcsin(\sin(b) \rightarrow a=b $

I know I'm doing something wrong, likely with the use of arcsin, but I'm not entirely sure.

Answer 1

$f$ is not one-to one:

$f(x)=f(x +2 \pi)$.