Exact solution for tan2x-cotx=0 in the interval [0,2pie)

[Westsiderider00]

find the exact solution of the equation tan2x-cotx=0 in the interval [0,2pie)
im having trouble solving this problem help would be nice plz.

 

[Westsiderider00]

I tried usingthe double angle formula and came up with :
(2tanx)/(1-tan2x)-1/tanx=0
but now i am stuck.....

 

[soroban]

Hello, Westsiderider00!

Edit: Sorry, I made a blunder . . . I'll correct it now.

Find the exact solution of: \(\displaystyle \,\tan2x\,-\,\cot x\:=\:0\) in the interval \(\displaystyle [0,\,2\pi)\)

We have: \(\displaystyle \,\tan2x\;=\;\cot x\;\;\Rightarrow\;\;\tan2x\;=\;\frac{1}{\tan x}\)

Double-angle formula: \(\displaystyle \,\frac{2\tan x}{1\,-\,\tan^2x} \;= \;\frac{1}{\tan x}\)

Clear denominators: \(\displaystyle \,2\tan^2x\;=\;1\,-\,\tan^2x\)

We have: \(\displaystyle \:3\tan^2x \:=\:1\;\;\Rightarrow\;\;\tan^2x\:=\;\frac{1}{3}\;\;\Rightarrow\;\;\tan x\:=\:\pm\frac{1}{\sqrt{3}}\)

Therefore: \(\displaystyle \,x\;=\;\frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{\7\pi}{6},\;\frac{11\pi}{6}\)

 

[Westsiderider00]

Hmm answer given in my book is different.....

Clear denominators: \,2\tan^2x\;=\;1\,-\,\tan x
How did you get a squared term getting rid of the denominators?

 

[Mrspi]

Hmm answer given in my book is different.....

Clear denominators: \,2\tan^2x\;=\;1\,-\,\tan x
How did you get a squared term getting rid of the denominators?

Multiply both sides of the equation by the common denominator of the fractions, which is (1 - tan² x)(tan x)

That will "clear" the denominators, since each divides into the multiplier, and leave you with this:

(2tan x)(tan x) = 1(1 - tan² x)