How do you find the domain of f(x) = tan(3arccos(x))?

Dec 15, 2017

$\text{By combining domain of arccos(x) and tan(x), see explanation:}$
$\left[- 1 , 1\right] \text{ \ {"-sqrt(3)/2", 0, "+sqrt(3)/2" }}$

Explanation:

$\text{arccos(x) is defined only for x "in" [-1, 1].}$
$\text{tan(x) is defined for all x values, except "pi/2 + k pi" , k integer.}$
$\text{So we check : }$
$\text{3 arccos(x) = "pi/2 + k pi".}$
=> arccos(x) = pi/6 + k pi/3"
=> x = cos(pi/6 + k pi/3)"
$\implies x = \pm \frac{\sqrt{3}}{2} , \mathmr{and} 0$
$\text{For those 3 x-values, the function f(x) is not defined, so we have}$
$\text{as domain :}$
$\left[- 1 , 1\right] \text{ \ {"-sqrt(3)/2", 0, "+sqrt(3)/2" }}$