Finding Domain of $\sqrt{sin(\sqrt{x})}$

[Yankel]

Hello all,

First, congratulations on the quick LATEX bar on the side, will try it now the first time.

I am looking for the domain of

$$\sqrt{sin(\sqrt{x})}$$

I need that the expression under the square root will be non-negative. The expression involves sin.

The sin function is non-negative at:

$$0\le\sqrt{x}\le\pi$$

$$2\pi\le\sqrt{x}\le3\pi$$

$$4\pi\le\sqrt{x}\le5\pi$$

And so on. From this I need to extract about x.

The answer for this, which I don't know where came from is:

$$4k^{2}\pi^{2}\le x\le(2k+1)^{2}\pi^{2}$$

How do I obtain the result from my information so far ?

Thanks !

 

[MarkFL]

I would begin by observing that we require:

\(\displaystyle 2k\pi\le\sqrt{x}\le(2k+1)\pi\) where \(\displaystyle k\in\mathbb{N_0}\)

and:

\(\displaystyle 0\le x\)

Now, what do you get when you square each part of the first inequality? Does the second still hold after doing so?

 

[Yankel]

it should hold.

if $$\sqrt{x}$$ is non-negative (and it is), then x also must be...