# What is the angle x if tan 3x -tan^3 x=0?

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Given the equation:

tan 3x - 2tan^3 x = 0

We need to find x values.

==> tan 3x = 2tan^3 x

Now from trigonometric identities we know that:

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

Let us substitute:

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

Now we will multiply by 1-3tan^2 x

==> 3tanx - tan^3 x = 2tan^3 x ( 1- 3tan^2 x)

==> 3tanx - tan^3 x = 2tan^3 x - 6tan^5 x

==> 6tan^5x - 3tan^3 x + 3tanx = 0

==> 3tanx(2tan^4 x - tan^2 x+1) = 0

==> Let u = tan^2 x

==> 3tanx ( 2u^2 - u +1) =0

But the equation between the brackets has no real solution.

Then the only solution is:

**tanx = 0 ==> x = 0, pi, 2pi **

**==> x = n*pi where n = 0, 1, 2, ....**

To find the angle x if tan 3x -2tan^3 x=0.

tan3x = (3tanx-tan^3x) / (1-3tan^2x) is an identity.

So we use this in the given equation:

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

(3tanx-tan^3x) - (1-3tan^2x)*2tan^3x = 0.

tanx {3 -tan^2x - 2tan^2 + 6tan^4x} = 0

tanx (3-3tan^2+6tan^4x) = 0. We divide by 3.

tanx (tan^4x-3tan^2x +1) = 0.

tanx(tan^2x-1)(1+2tan^2) = 0.

tanx = 0, tan^2-1 = 0, or 1+2tan^2x = 0.

tanx = 0, ortan^2 =1, or tan^2 = 1. So tanx = +or- 1.

So x = npi, or x= npi+pi/4, or npi - pi/4, for n = 0,1,2,..

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.**