Two roots of $\arcsin(x)$ in the range $[0,2 \pi]$

Question

I am baffled with how to write the two roots of arcSin$(x)$ in the range $[0,2 \pi]$, while $x \in [-1,1]$, such that one root can be directly calculated in terms of the other root.

For instance, we can write the two roots of arcCos$(x)$ in the range $[0,2 \pi]$,while $x \in [-1,1]$, as $\theta_1=\arccos(x)$, and $\theta_2= 2 \pi - \theta_1 $.

But how to do that for $\arccos(x)$ ?

Answer 1

If $x >0$, then the two roots of $\sin\theta = x$ satisfy $\theta_1 + \theta_2 = \pi$. If $x<0$, then the two roots satisfy $\theta_1+\theta_2 = 3\pi$.

And to correct what you wrote about $\cos\theta = x$, the two roots satisfy $\theta_1 + \theta_2 = 2\pi$.