Question 787558
{{{3tan(x)-sqrt(3)=0}}}
{{{3tan(x)=sqrt(3)}}}
{{{tan(x)=sqrt(3)/3}}}
 
We know that {{{tan(pi/6)=sqrt(3)/3}}} (or {{{tan(30^o)=sqrt(3)/3}}} if using degrees). For that angle {{{sin(pi/6)=1/2}}} and {{{cos(pi/6)=sqrt(3)/2}}}
That is the only solution for {{{-pi/2<x<pi/2}}} (or {{{-90^o<x<90^o}}} if using degrees).
The function does not exist for {{{x=-pi/2}}} or {{{x=pi/2}}}, where {{{cos(x)=0}}}.
In between those values {{{tan(x)}}} increases continuously.
 
We also know that if {{{tan(x)}}} exists, {{{tan(x+pi)=tan(x)}}}, so the funcion's values repeat at {{{pi}}} intervals, and we say that the function has period {{{pi}}}.
So there are infinite solutions that can be all expressed as
{{{highlight(x=pi/6+k*pi)}}} for all {{{k}}} integers (or {{{highlight(x=30^o+k*180^o)}}} for all {{{k}}} integers if you measure angles in degrees).