# How do you find the domain and range for y = 5arcsin(2cos(3x))?

Jun 11, 2015

"Domain"={x|x in RR ^^( pi/9+2/3kpi<=x<=2/9pi+2/3kpi vv 4/9pi+2/3kpi<=x<=5/9pi+2/3kpi) " where "kinZZ}
$\text{Range} = \left\{y | y \in \mathbb{R} \wedge - \frac{5}{2} \pi \le y \le \frac{5}{2} \pi\right\}$

#### Explanation:

$\arcsin \left(x\right) : \left[- 1 , 1\right] \to \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
So the argument of the function must be between 1 and -1.
$- 1 \le 2 \cos \left(3 x\right) \le 1$
So we have to solve the system:
$\left\{\begin{matrix}2 \cos \left(3 x\right) \ge - 1 \\ 2 \cos \left(3 x\right) \le 1\end{matrix}\right.$
$\left\{\begin{matrix}\cos \left(3 x\right) \ge - \frac{1}{2} \\ \cos \left(3 x\right) \le \frac{1}{2}\end{matrix}\right.$
$\frac{\pi}{3} + 2 k \pi \le 3 x \le \frac{2}{3} \pi + 2 k \pi \vee \frac{4}{3} \pi + 2 k \pi \le 3 x \le \frac{5}{3} \pi + 2 k \pi$
So, isolating x, the domain would be:
$\frac{\pi}{9} + \frac{2}{3} k \pi \le x \le \frac{2}{9} \pi + \frac{2}{3} k \pi \vee \frac{4}{9} \pi + \frac{2}{3} k \pi \le x \le \frac{5}{9} \pi + \frac{2}{3} k \pi$.

Considering that $2 \cos \left(3 x\right)$ in the domain would vary in $\left[- 1 , 1\right]$, the range of $\arcsin \left(x\right)$ willl vary in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$. But the function has a cohefficient "5" that extends the amplitude of our function to $- \frac{5}{2} \pi , \frac{5}{2} \pi$.

This strange function would be something like this: 