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.


2 Answers 2


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


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

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


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