How do i figure out which period the function is in?
Im doing the problem 3 tan^{2 }x-1=0
i got tan x= 1sqrt 3 i have to find a solution to x. How do i know tan is on the interval 0, pi and not 0, 2pi?
The period of a periodic function $\displaystyle f(x)$ is the smallest positive real number $\displaystyle \alpha$ such that $\displaystyle f(x+\alpha)=f(x)$ for all $\displaystyle x$ in the domain of the function. In case of $\displaystyle \tan,$ $\displaystyle \tan(x+\pi)=\tan x$ for all real $\displaystyle x$ and no positive real number $\displaystyle \alpha$ smaller than $\displaystyle \pi$ satisfies $\displaystyle \tan(x+\alpha)=\tan x$ for all $\displaystyle x.$ (It's true that $\displaystyle \tan(x+2\pi)=\tan x$ for all $\displaystyle x$ but $\displaystyle 2\pi$ is not the smallest such positive real number.)
$\displaystyle 3\tan^2{x} - 1 = 0$
$\displaystyle \tan{x} = \pm \frac{1}{\sqrt{3}}$
since the solution interval is not given, then all possible solutions are ...
$\displaystyle x = \frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$
$\displaystyle x = -\frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$