Domain of $f(x) = \sqrt{\sin(\sqrt{x})}$

Question

I have a question regarding the domain of $f(x)$ when equal to the root of the root of sinus. In other words,

$f(x) = \sqrt{\sin (\sqrt{x})}$.

What I do know is that the domain of $\sqrt{\sin}$ must be larger than $0$, which means is between $[0,\pi]$. However that's clearly only half of the problem.

As much as I try to, I can't wrap my head around this. Could anyone help?

Answer 1

First condition: $x$ must be nonnegative.

Second condition:$x$ must satisfy $$\sin\sqrt x \ge 0\iff 2k\pi\le \sqrt x\le \pi+2k\pi=(2k+1)\pi. $$

Can you proceed?

Answer 2

Work this from inside out.

The domain of $\sqrt x$ is $x\ge0$. The sine has no domain limitation. Then the outer square root needs a non-negative argument, which occurs when the argument of the sine is in range $[0,\pi]$ plus $2k\pi$ (due to periodicity).

Hence

$$\sqrt x\in[2k\pi,(2k+1)\pi]$$

or

$$x\in[(2k\pi)^2,((2k+1)\pi)^2].$$

To avoid duplicates, $k\ge0$.