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

one-one Many-one Bijective None of these

Many-one

$f(x)=\sin x,\,x∈R$ 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.