Question 148708
let tan x= u

3u^3=u
3u^3-u=0
u(3u^2-1)=0
so u=0 or 3u^2=1 which is the same as u=+/- 1/sqrt(3)

So we want
tan x =0
and
tan x=sinx/cosx=+/- sqrt(3)/3

tan x is 0 when sine is 0. That occurs at 0 degrees and 180 degrees, or 0 radians and pi radians.

The positive value for 1/sqrt(3) occurs at cos x=sqrt(3)/2 and sin x=1/2, AND cos x=-sqrt(3/2) and sinx=-1/2. Which is to say, 30 degrees or pi/6 radians AND 210 degrees or 7pi/6.

The negative value occurs at cos x=sqrt(3)/2 and sin x=-1/2 ----- or cos x=-sqrt(3)/2 and sin x=1/2. Respectively, these occur at 330 degrees which is 11pi/6, and at 150 degrees which is 5pi/6.



So our entire solution set is:
x=0, x=pi/6, x=5pi/6, x=pi, x=7pi/6, x=11pi/6.