The following condition fails to define a function on any domain. State. why?" $\sin f(x) =x$ 
Quoting" The following condition fails to define a function on any domain. State. why?"
$$\sin f(x) =x$$

I understand that it yields a sinusoid oscillating vertically defined between $x=-1$ and $x=1$.
It shows that $\forall x \in [-1,1]$, many values of y is mapped onto.
Because there is no unique image for a specific value of $x$, it fails to define a function.
Is there a more formal approach more appropriate for a real analysis class? or is there an alternative explanation?
Any input is much appreciated.
 A: ALTERNATIVE EXPLANATION: The reason that this does not define a function is because it fails the most elementary of all function tests: the "vertical line" test. It even defies the very definiton of "function"! If this function did exist, here is what it would look like:

For all $x$ between $-1$ and $1$, the function would have infinitely values, and then would be undefined for all other values! This cannot be! The definition of a function even states that it cannot have more than one output value for each input.
A: If $|x|\leq 1$ the statement $\sin f(x)=x$ does not DEFINE $f(x)$ although the statement may be a property of a function $f.$  There are infinitely many $y$ for which $\sin y=x$ and the statement $\sin f(x)=x$ does not tell you which one of these $y$ is actually $f(x).$
A: If a graph is a function then it passes the vertical line test ie the graph cuts a given vertical line atmost $1$ time. But as here our graph is of $\arcsin (x) $ which ust oscillates between $\frac {-\pi}{2},\frac {\pi}{2} $. Hence the given condition isnt a function on any domain.
