Question 1206427

{{{tan^2 (x) + tan(x)=0 }}}

{{{tan (x) (tan(x)+1)=0 }}}

solutions:

{{{tan(x)=0}}}
or
{{{tan(x)=-1}}}


then 

{{{x=tan^-1(0)}}}
{{{x=0}}}

or

{{{x=tan^-1(-1)}}}

{{{x=-pi/4}}}


in the interval [{{{0}}},{{{2pi}}})

a function {{{tan(x)}}} is a periodic function and has a period of {{{ pi}}} 

{{{x=0}}}
{{{x=0+pi=pi}}}
{{{x=-pi/4+pi=3pi/4}}}
{{{x=3pi/4+pi=7pi/4}}}