Check the nature of the following function. f(x)=sin x, x∈R

$f(x)=\sin x,\,x∈R$

Let $f(x_1)=f(x_2)$

$⇒\sin x_1=\sin x_2$

$⇒ x_1=nπ+(-1)^nx_2,\,n∈Z$

There are infinite pair $(x_1,x_2),x_1≠x_2$ which satisfy above equation.

i.e., $f(0)=f(π)=0$

So, f(x) is many-one.