Question

Find all solutions in the interval [0, 2π).

7 tan3x - 21 tan x = 0

Answer 1

For the answer to the question above, I will make a step by step on how to solve the problem
7 tan^3x - 21 tanx = 0
=> 7tanx (tan^2 x - 3) = 0
=> tanx = 0 or tan^2 x = 3
=> x = 0 or π
or tanx = ± √3, i.e., tanx = √3 or tanx = - √3
tanx = √3
=> tanx = tan (π/6) or tan (π + π/6)
=> x = π/6 or 7π/6
and
tanx = - √3
=> tanx = tan (π - π/6) or tan (2π - π/6)
=> x = 5π/6 or 11π/6

The answer would be x = { 0, π/6, 5π/6, π, 7π/6, 11π/6 }.
I hope my answer helped you.

Answer 2

Hence, all the solutions in the interval [0,2π) are:

`0\ ,\pi\ ,(\pi)/(3)\ ,(2\pi)/(3)\ ,(4\pi)/(3)\ ,(5\pi)/(3)`

Step-by-step explanation:

We are asked to find the solution of the trignometric identity which is given by:

`7\tan^3x-21\tan x=0`

On dividing both side by 7 we get:

`\tan^3x-3\tanx=0\\\\i.e.\\\\\tan x(\tan^2 x-3)=0`

i.e.

Either

`\tan x=0`

i.e.

`x=0,\pi`

or

`\tan^2x-3=0\\\\i.e.\\\\\tan^2x=3\\\\i.e.\\\\\tan x=\pm √(3)`

If

`\tan x=√(3)\\\\Then\\\\x=(\pi)/(3),(4\pi)/(3)`

and if

`\tan x=-√(3)\\\\Then\\\\x=\pi-(\pi)/(3)=(2\pi)/(3)\\\\and\\\\x=2\pi-(\pi)/(3)\\\\i.e.\\\\x=(5\pi)/(3)`

Hence, all solutions are:

`0\ ,\pi\ ,(\pi)/(3)\ ,(2\pi)/(3)\ ,(4\pi)/(3)\ ,(5\pi)/(3)`