# What is the solution(s) to the following, given that the exact solutions for x lie in `[0, 2pi)`  ?  `tan^3(x) = 1/3 tanx`

ishpiro | Certified Educator

`tan^3x =1/3 tanx`

Bring 1/3 tanx to left side:

`tan^3x - 1/3tanx = 0`

Factor out one tanx:

`tanx(tan^2x - 1/3) = 0`

Now the equation breaks up into two:

tanx = 0 and `tan^2x -1/3 = 0`

The first equation has only solution x = 0 on the interval `[0, 2pi)`

The second equation can be rewritten by adding 1/3 to both sides:

`tan^2x = 1/3`

Take square root of both sides:

`tanx = +-1/sqrt3 = +-sqrt3/3`

The tangent achieves the value of `sqrt3/3` when `x= pi/6` and `x=(7pi)/6`.

The tangent achieves the value of `-sqrt3/3`  when `x = (5pi)/6`  and `x=(11pi)/6`  .

Therefore, these are the solutions of the given equation on `[0, 2pi)` :

`x = 0, pi/6, 5pi/6, 7pi/6, 11pi/6`

oldnick | Student

`tan^3x=1/3 tan x`

dividing by tanx x:

`tanx=0`    `x=kpi`

`tan^2x=1/3`     `rArr`  `tanx=+- sqrt(3)/3`   `x= pi/6+ pi`  `x=5/6 pi +kpi`

So the required solutions are:

`x=kpi;x=pi/6+kpi;x=5/6pi+kpi`