# 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`

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`tan^3x =1/3 tanx`

Bring 1/3 tan*x* to left side:

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

Factor out one tan*x*:

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

Now the equation breaks up into two:

tan*x* = 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` **

`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`