domain and range of trigonometric function I have a question relating to finding domain and range.
If you have something like $f(x) = \sqrt{\sin x}$, or $\sqrt{\sin 3x}$, how to you work out the domain for this. I have viewed an answer to see if I could work it out myself, but I am struggling to understand it. I realise this function is an indefinite disjointed interval, however I struggling to understand how to express the interval when the endpoints are fixed and infinite??
The domain is expressed as the following:
$0 \leq 3x - 2\pi n \leq \pi$ and $n\in\mathbb{Z}$.
I am unsure where the n has come from? Also, how is it determining the regular value pie. Is $\mathbb{Z}$ referring to imaginary numbers. I am only just getting into mathematics again. Thank you.
 A: So, you have $f(x) = \sqrt {\sin(x)}$, and you want to find the domain. The domain would be the range of values for which $f$ is defined. Well, where is $f$ defined? It's a composition of functions, so consider:


*

*$\sin(x)$ is defined for all real numbers $x$

*$\sqrt x$ is defined for all nonnegative reals $x$
(Based on your intended answer, I assume you want just real numbers. Technically you can define the above for complex numbers but that feels beyond the scope of your question.)
So, $f$ is undefined, wherever $\sin(x) < 0$. Where is $\sin(x) < 0$ then? It becomes quite obvious if you look at a graph of it: on the intervals...


*

*$(\pi, 2\pi)$

*$(3\pi, 4\pi)$

*$(5\pi, 6\pi)$

*$(7\pi, 8\pi)$
...and so on and so forth. And the same is true going backwards as well. This means, in general, $\sin(x) < 0$ on all intervals $((2k+1)\pi, 2k\pi)$ where $k$ is an integer: you can see this just by pattern-matching the various intervals and constructing this form.
In turn, this means $f$ is defined on the intervals $[0, \pi]$ and $[2\pi, 3\pi]$ and $[4 \pi, 5\pi]$ and ... so on and so forth, i.e. the intervals of the form $[2k \pi, (2k+1) \pi ]$ for $k \in \Bbb Z$.
This is a perfectly acceptable way of writing the domain of $f$ as described above: or you could write it as a union:
$$\bigcup_{k \in \Bbb Z} [2k \pi, (2k+1) \pi ]$$
If you insist on writing it as an inequality, consider what each interval means for $x$: namely, $2k \pi \le x \le (2k+1) \pi$. Subtract $2k \pi$ throughout and simplify to get $0 \le x - 2k\pi \le \pi$.

This all of course solves the first of your problems, but the same idea applies to solving the second.
A: Domain is is the X-axis where $x\in\mathbb{R}$, i,e. $-\infty \le x \le \infty$. Range is the Y-Axis and, here, $-1\le\sin x \le1$. For your function(s), the value of$f(x)$ changes as rapidly as the coefficient of sine and, if $\sin x\lt 0,  f(x)\le i, \text { where }i=\sqrt{-1}$. So your range is two-fold. Be careful not to mix the ranges because it is not possible to "order" complex numbers. Only purely real $(\sin x \ge 0)$ or purely imaginary numbers $(\sin x \lt 0)$ can be ordered.
On the other hand, if imaginaries are not allowed, your domain is
$$0\le3x-2\pi n\le\pi,\quad n\in\mathbb{Z} \qquad\implies\qquad0\le x\le\frac{(2n +1)\pi}{3},\quad  n\in\mathbb{Z}$$
Now, with $x$ shown in a different light, we an see that the domain is not continuous but rather $1/3$ fractional multiples of $\pi$.  These values correspond to $60^\circ$, $120^\circ$, $180^\circ$, ... and the domain is limited to 3 distinct values
$$\sin x\in\bigg\{-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\bigg\}$$
Draw a picture and it will make sense.
A: Basically, we have to solve $\sin(\omega x)\geq0$, because we know that $f(x)=\sqrt{\sin(\omega x)}$ is defined when the argment of the square root is positive.
So, we have:
$$\sin(\omega x)\geq0$$
First, we have to solve:
$$\sin(\omega x)=0\leftrightarrow \omega x=0+k\pi, k \in Z \leftrightarrow x=\frac{k}{\omega}\pi$$
In $[0,2\pi]$, we know that $\sin(x)\geq0$ when $0\leq x \leq \pi$, and so generalizing:
$$\frac{2k}{\omega}\pi \leq x \leq \frac{2k}{\omega}\pi+\frac{\pi}{\omega}$$
