Find the domain of $$\sqrt{\sin\pi(x^2-x)}$$

I'm confused about this function. I've been trying to figure out what's the domain but can't get a right answer.

  • $\begingroup$ In other words, please include your thoughts and efforts in this and future posts. Formatting tips here. $\endgroup$
    – Em.
    Commented Jan 14, 2016 at 0:18
  • 2
    $\begingroup$ I would write that as $\sqrt{\sin(\pi(x^2-x))}$ $\endgroup$
    – Motun
    Commented Jan 14, 2016 at 0:28
  • $\begingroup$ Find the values of $x$ that make the $\sin$ positive only. $\endgroup$ Commented Jan 14, 2016 at 0:41
  • 1
    $\begingroup$ @ja2, or possibly zero. $\endgroup$
    – JRN
    Commented Jan 14, 2016 at 0:58

3 Answers 3


Start with the inequality


To understand when this is true let's look at the inequality


It is clear that

$$2\pi n\le t\le 2\pi n +\pi, \quad n\in\mathbb Z$$

Substituting $\pi(x^2-x)$ for $t$ we see

$$2\pi n \le \pi(x^2-x) \le \pi (2n +1),\quad n\in\mathbb Z$$

$$2n\le x^2-x\le 2n+1, \quad n\in\mathbb Z$$

This can be interpreted geometrically as bounding a parabola by 2 horizontal lines at $y=2n+1$ and $y=2n$ which means we have to deal with the left and right solutions.


From this image it is also clear that there is no solution when $n<0$

Let's solve the lower line first. We have

$$x^2-x\ge 2n$$

$$x\ge\frac 1 2 \left(1+\sqrt{8n+1}\right)\quad\text{or}\quad x\le\frac 1 2 \left(1-\sqrt{8n+1}\right)$$ For $x^2-x\le 2n+1$ it can be shown that

$$x\le \frac 1 2 \left(1+\sqrt{8n+5}\right)\quad\text{or}\quad x\ge \frac 1 2 \left(1-\sqrt{8n+5}\right)$$

Combining these inequalities we obtain

$$\frac 1 2 \left(1+\sqrt{8n+1}\right)\le x\le \frac 1 2 \left(1+\sqrt{8n+5}\right)\quad\text{or}\quad \frac 1 2 \left(1-\sqrt{8n+5}\right)\le x\le \frac 1 2 \left(1-\sqrt{8n+1}\right), \quad n\in\mathbb Z,n\ge 0$$


We know that $\sqrt{x} \geq 0$ and so $\sin{\pi(x^2-x)} \geq 0$. The function $\sin{x}$ is nonnegative on $\ldots,[-2\pi,-\pi],[0,\pi], [2\pi, 3\pi],\ldots$. Thus, $x^2-x$ must be in $\ldots,[-2,-1], [0,1], [1,2] \ldots$. So take the union of all $x$ such that $x^2-x$ is in each interval and that is the domain.

In more mathematical terms, $x^2-x \in \displaystyle \bigcup_{i\in\mathbf Z} [2i,2i+1]$ and thus $$x \in \displaystyle \bigcup_{i\in\mathbf Z}\left [\dfrac{1}{2}(1 + \sqrt{8i+1}), \dfrac{1}{2}( \sqrt{8i+5})\right ] \bigcup \bigcup_{i\in\mathbf Z}\left[\dfrac{1}{2}(1 - \sqrt{8i+5}),\dfrac{1}{2}(1 - \sqrt{8i+1}) \right].$$



$p(x)=x^2-x$ must lie in $[2k,2k+1]$ for some $k\in\mathbf Z$. Hence you have to determine $$\displaystyle\bigcup_{k\in\mathbf Z}p^{-1}([2k,2k+1]).$$ Sketching the function may help.

  • $\begingroup$ Why $p^{-1}$?...Thx. $\endgroup$
    – NoChance
    Commented Jan 14, 2016 at 0:32
  • $\begingroup$ By definition, $p^{-1}([2k,2k+1])=\{x\mid p(x)\in [2k,2k+1]\}$. It amounts to solving quadratic inequations in pairs. $\endgroup$
    – Bernard
    Commented Jan 14, 2016 at 0:35
  • $\begingroup$ Thank you for your explanation. $\endgroup$
    – NoChance
    Commented Jan 15, 2016 at 4:39

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