Help solving trig equation

[Stormin63]

Hi,

I'm having trouble solving the following for 0 > x > 2pi

tan x = 2 + tan 3x

i've used tan 3x = tan (x + 2x) = (tan x)(tan 2x) / 1 - (tan x)(tan 2x) a couple of times to get the equation in terms of tan x,

I.e.

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

which may or may not be correct.

but I'm not sure how to solve this polynomial.

Any help is much appreciated!

p.s. From Wolfram Alpha the solutions are pi*n - 3pi/4, pi*n - 5pi/8 and pi*n - pi/8

 

[Deleted member 4993]

Hi,

I'm having trouble solving the following for 0 > x > 2pi

tan x = 2 + tan 3x

i've used tan 3x = tan (x + 2x) = (tan x)(tan 2x) / 1 - (tan x)(tan 2x) a couple of times to get the equation in terms of tan x,

I.e.

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

which may or may not be correct.

but I'm not sure how to solve this polynomial.

Any help is much appreciated!

p.s. From Wolfram Alpha the solutions are pi*n - 3pi/4, pi*n - 5pi/8 and pi*n - pi/8

Substitute

tan(x) = u

then

3u^3 - 4u^2 - u +2 = 0

By observation, u = 1 is a solution. Then by dividing the polynomial by (u-1) we get

(u - 1)(3u^2 - u -2) = 0

Now factorize the quadratic and continue....