Trig periods

**noork85** · May 17th 2012, 10:57 AM

How do i figure out which period the function is in?
Im doing the problem 3 tan²x-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?

**Sylvia104** · May 17th 2012, 12:26 PM

The period of a periodic function $f(x)$ is the smallest positive real number $\alpha$ such that $f(x+\alpha)=f(x)$ for all $x$ in the domain of the function. In case of $\tan,$ $\tan(x+\pi)=\tan x$ for all real $x$ and no positive real number $\alpha$ smaller than $\pi$ satisfies $\tan(x+\alpha)=\tan x$ for all $x.$ (It's true that $\tan(x+2\pi)=\tan x$ for all $x$ but $2\pi$ is not the smallest such positive real number.)

**skeeter** · May 17th 2012, 02:00 PM

Originally Posted by noork85

How do i figure out which period the function is in?
Im doing the problem 3 tan²x-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?

$3\tan^2{x} - 1 = 0$

$\tan{x} = \pm \frac{1}{\sqrt{3}}$

since the solution interval is not given, then all possible solutions are ...

$x = \frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$

$x = -\frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$

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