How to find roots of $\sin (x) - a$?

Question

How to find roots of $\sin(x) - a$, where $a \in [0, 1)$ and $x \in [0, 2\pi]$?

Answer 1

The roots are $\arcsin(a)$ and $\pi - \arcsin(a)$.

Answer 2

The inverse function for $\sin $ is called arcsinus and noted $\arcsin x$. More on this function here.

Answer 3

$$x=k\pi+(-1)^k\sin^{-1}{(a)}$$ so when $k$ is even, it is $k\pi+\sin^{-1}{(a)}$, and when $k$ is odd, it is $k\pi-\sin^{-1}{(a)}$