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

Answer:

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

$0\ ,\pi\ ,\dfrac{\pi}{3}\ ,\dfrac{2\pi}{3}\ ,\dfrac{4\pi}{3}\ ,\dfrac{5\pi}{3}$

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

$\tan^2x-3=0\\\\i.e.\\\\\tan^2x=3\\\\i.e.\\\\\tan x=\pm \sqrt{3}$

If

$\tan x=\sqrt{3}\\\\Then\\\\x=\dfrac{\pi}{3},\dfrac{4\pi}{3}$

and if

$\tan x=-\sqrt{3}\\\\Then\\\\x=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\\\\and\\\\x=2\pi-\dfrac{\pi}{3}\\\\i.e.\\\\x=\dfrac{5\pi}{3}$

Hence, all solutions are:

            $0\ ,\pi\ ,\dfrac{\pi}{3}\ ,\dfrac{2\pi}{3}\ ,\dfrac{4\pi}{3}\ ,\dfrac{5\pi}{3}$