Question 371832
{{{3(tanx)^3 - tanx = 0}}},
{{{tanx(3(tanx)^2 - 1) = 0}}},
tanx = 0, or {{{(tanx)^2 = 1/3}}}, or
tanx = 0, or {{{tanx = 1/sqrt(3)}}} or {{{tanx = -1/sqrt(3)}}}. 
 therefore
x = 0, {{{pi}}}, {{{pi/6}}}, {{{7pi/6}}}, {{{5pi/6}}}, {{{11pi/6}}}.