Find the range of arcsin$((1-x^2)^{0.5})$

Question

Title says it all, how do you get the answer to this?

So far I only reach $0<1-x^2<pi/2$ but I get an invalid answer from here. the correct answer is $0<x<pi/2$.

Any help is appreciated, thanks.

Answer 1

Hint: the function $\sqrt{1-x^2}$ takes its values on $[0,1]$; what are $\arcsin0$ and $\arcsin1$? And what can you say about the arcsine function? Consider the most fundamental properties.

Answer 2

$$1-x^2\le1$$ $$0\le\sqrt{1-x^2}\le1$$ $$-\frac\pi2\le\sin^{-1}y\le\frac\pi2$$ If $0\le y\le1$, what values can $\sin^{-1}y$ take? Where is sine positive?