# Domain of $g(x)=\sqrt{\sin(π\sin(πx))}$

Let a function $$g(x)=\sqrt{\sin(π\sin(πx))}$$

If I have want to find its domain. The first step wil be making the term inside the square root greater than or equal to zero i.e.
$$\sin(π\sin(πx)) \ge 0$$

Further we can say $$2nπ \le π\sin(πx) \le (2n+1)π$$

Where $$n \in Z$$

$$2n \le \sin(πx) \le (2n+1)$$

But now I am stuck. Further I think here we will be using inverse trigonometry but I have not been familiar with inequalities involving inverse trigonometry.

So far you made good progress, you are not missing much.

After simplifying the $$\pi$$ you get $$2n\leq \sin(\pi x)\leq 2n+1$$ now, knowing that $$\sin$$ ranges in $$[-1,1]$$ you only have to consider $$n=0$$ and the special case $$\sin(\pi x)=-1$$ (with $$n=-1$$).

Thus you end up with the case $$0\leq \sin(\pi x)\leq 1$$ which as before leads to $$2m\pi\leq \pi x\leq (2m+1)\pi$$ thus $$2m\leq x\leq (2m+1)$$

and the special case is just $$x=-\frac{1}{2}+2m$$

This is confirmed by WA

Also

Graph by Desmos

You can use Mathematica to calculate the result.

n = 2;
f = Nest[Sin[Pi*#] &, x, n]
Reduce[f >= 0, x, Reals] // FullSimplify


$$c_1\in \mathbb{Z}\land (2 x+1=4 c_1\lor 2 x=3+4 c_1\lor 0\leq x-2 c_1\leq 1)$$