# Check wheather it is possibleCheck if there is an angle x so that tan 3x = tan^3 x?

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You need to write `tan 3x` in terms of `tan x` , using trigonometric identity, such that:

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

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

Reducing duplicate factors yields:

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

Replacing `(3tan x - tan^3 x)/(1 - 3tan^2 x)` for `tan 3x` in equation yields:

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

You need to factor out `tan x` such that:

`tan x((3 - tan^2 x)/(1 - 3tan^2 x) - tan^2 x) = 0 => {(tan x = 0),((3 - tan^2 x)/(1 - 3tan^2 x) - tan^2 x = 0):}`

`tan x = 0 => x = npi`

`3 - tan^2 x - tan^2x + 3tan^4 x = 0`

`3tan^4 x - 2tan^2 x + 3 = 0`

Replacing t for `tan^2 x` yields:

`3t^2 - 2t + 3 = 0 => t_(1,2) = (2+-sqrt(4 - 36))/6 `

`t_(1,2) = (2+-sqrt(-32))/6 => t_(1,2) = (2+-4i*sqrt2)/6`

`tan^2 x = (2+-4i*sqrt2)/6`

Since the range of tangent function is the real set of numbers, hence, `tan^2 x = (2+-4i*sqrt2)/6` are invalid solutions.

**Hence, evaluating the solution to trigonometric equation yields **`x in {n*pi | n in Z}.`

We'll write tan 3x = tan (2x + x)

tan 3x = (tan 2x + tan x)/(1 - tan 2x*tan x)

tan 2x = 2tan x/[1 - (tanx)^2]

tan 3x = 3tanx - (tan x)^3]/[1 - 3(tanx)^2]

we'll re-write the equation:

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

3tanx - (tan x)^3 - (tan x)^3 + 3(tan x)^5 = 0

We'll eliminate like terms:

3tanx + 3(tan x)^5 = 0

We'll factorize by 3tan x:

3tan x[1 + (tan x)^4] = 0

We'll set each factor as zero:

3tan x = 0

tan x = 0

x = k*pi

1 + (tan x)^4 = 0

(tan x)^4 = -1

(tan x)^2 = i impossible

(tan x)^2 = -i imposible

**The solution of the equation is x = k*pi.**