Question

How do you solve `tan x + tan 3x = 0` ?

Answer 1

Solutions:

`x = npi" "` or `" "x = +-pi/4+npi`

for any integer `n`

Explanation:

Formula for `tan 3 theta`

By de Moivre's formula we have:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

In particular:

`cos 3 theta + i sin 3 theta =(cos theta + i sin theta)^3`

`color(white)(cos 3 theta + i sin 3 theta) =cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin^2 theta - i sin^3 theta`

`color(white)(cos 3 theta + i sin 3 theta) =(cos^3 theta- 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)`

So equating real and imaginary parts, we find:

`{ (cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta), (sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta) :}`

So:

`tan 3 theta = (sin 3 theta) / (cos 3 theta)`

`color(white)(tan 3 theta) = (3 cos^2 theta sin theta - sin^3 theta) / (cos^3 theta - 3 cos theta sin^2 theta)`

`color(white)(tan 3 theta) = ((3 cos^2 theta sin theta - sin^3 theta) -: cos^3 theta) / ((cos^3 theta - 3 cos theta sin^2 theta) -: cos^3 theta)`

`color(white)(tan 3 theta) = (3 tan theta - tan^3 theta) / (1 - 3 tan^2 theta)`

Example

So:

`0 = tan x + tan 3 x`

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

Multiplying both ends by `(1-3 tan^2 x)` this becomes:

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

`color(white)(0) = tan x - 3 tan^3 x + 3 tan x - tan^3 x`

`color(white)(0) = 4 tan x - 4 tan^3 x`

`color(white)(0) = 4 tan x(1 - tan^2 x)`

`color(white)(0) = 4 tan x(1 - tan x)(1 + tan x)`

Hence:

`tan x = 0`, `1` or `-1`

Note that `tan x` has period `pi`, so all the solutions are:

`x = tan^(-1) (0) + npi = npi`

`x = tan^(-1) (1) + npi = pi/4+npi`

`x = tan^(-1) (-1) + npi = -pi/4+npi`

for any integer `n`