Why does $\sin{x} = \frac{1}{2}$ have two solutions but $\arcsin{\frac{1}{2}}$ has one solution? Can anyone help me understand why $\sin{x} = \frac{1}{2}$ have two solutions but $\arcsin{\frac{1}{2}}$ has one solution?
Aren't they equivalent? 
 A: It is the same reason that
$$
x^2 = 1
$$
has two solutions ($x = 1$ and $x = -1$) but $\sqrt{1}$ has only one solution (the square root of $1$ is $1$, not $-1$.)
The weird thing about $\arcsin y$ is that it does NOT give you all possible values $x$ such that $\sin x = y$. It cannot do that, because we want it to be a function, in other words, for every input there is only exactly one output. So if I put in $\frac12$ for $y$ and look at $\arcsin \frac12$, $\arcsin$ can only give me one output, not multiple outputs. So $\arcsin$ gives me back $\frac{\pi}{6}$, even though there are infinitely other many values of $x$ that satisfy the equation. $\arcsin$ just gives me ONE of them.
In summary, there are infinitely many $x$ such that $\sin x = \frac12$, but $\arcsin$ is a function so it can only give one of them back. $\arcsin \frac12$ is therefore just A solution to the equation, not ALL solutions.
A: Since the sine function is not one-to-one, the sine function does not have an inverse. Arcsine (or "Inverse sine"), is actually the inverse of the restricted sine function from $-\frac{\pi}2$ (radians) to $\frac{\pi}2$ (radians). Therefore, the arcsine of $\frac{1}2$ only has on solution on that particular interval, but since the (regular) sine function is periodic, the first equation you asked about has not only 2 solutions, but infinate solutions.
