Solve the equation on the interval [ 0 , 2 π ] tan ⁡ 2...

We then divide $\tan 2x$ and $\tan x$ by what has been taken out or $\tan x$.
After factoring, we get this trig equation in factored form:
$\tan x(\tan x-1)=0$
We have two factors and they are:
$\tan x$ and $\tan x-1$
Set each factor to zero just like you did in algebra 1 andsolve for the given trig function.
$\tan x=0$
When does $\tan x=0$?
$\tan x=0$ at 0 degrees (the same as 360 degrees) and 180 degrees.
We now set the other factor to 0 and solve the equation.
$\tan x-1=0$
Adding 1 to both sides we get:
$\tan x=1$
When does $\tan x=1$?
$\tan x=1$ at 45 degrees and 225 degrees
All degrees between the given interval are: 0 degrees (or 360degrees), 180 degrees, 45 degrees and 225 degrees.
If you want the answer in radians, here is it:
All radian measures are:
$\frac{\pi }{4},\pi ,\frac{5\pi }{4}$, and $2\pi$

$\tan 2x=\frac{2\tan (x)}{1-{\tan }^2(x)}$
$\frac{2\tan (x)}{1-{\tan }^2(x)}-\frac{\tan (x)}{1-{\tan }^2(x)}=0$
$\frac{2\tan (x)-\tan (x)(1-{\tan }^2(x))}{1-{\tan }^2(x)}=0$
$2\tan (x)-\tan (x)+\tan (x){\tan }^2(x)=0$
$\tan (x)+\tan (x){\tan }^2(x)=0$
$\tan (x)(1+{\tan }^2(x))=0$
$\tan (x)=0$ or ${\tan }^2(x)=-1$
x = 0 or $2\pi$