Find the domain of f(x)=sin(x)f(x) = \sqrt{\sin(\sqrt{x})}f(x)=sin(x).
[0,π2][0, \pi^2][0,π2]
⋃k=0∞[(2kπ)2,((2k+1)π)2]\bigcup_{k=0}^{\infty} [(2k\pi)^2, ((2k+1)\pi)^2]⋃k=0∞[(2kπ)2,((2k+1)π)2]
[0,2π][0, 2\pi][0,2π]
R+\mathbb{R}^+R+