Question

How do you find the exact solutions of the equation `tan2x-cotx=0` in the interval `[0,2pi)`?

Answer 1

`pi/6, pi/2`

Explanation:

tan 2x - cot x = 0
`(sin 2x)/(cos 2x) - cos x/(sin x) = 0`
`(sin x.sin 2x - cos x.cos 2x)/(sin x.cos 2x) = 0`
(sin x.sin 2x - cos x.cos 2x) = 0
Use trig identity: cos (a + b) = cos a.cos b - sin a.sin b
In this case we have:
sin x.cos 2x - cos x.cos 2x = - cos (x + 2x) = - cos 3x = 0
cos 3x = 0
Trig unit circle gives:
a. `3x = pi/2` --> `x = pi/6`
b. `3x = (3pi)/2` --> `x = (3pi)/6 = pi/2`
Answers for `(0, 2pi)`: `pi/6 and pi/2`
Check:
If `x = pi/6` --> `2x = pi/3` --> `tan 2x = sqrt3` --> `cot x = cot (pi/6) = sqrt3`, then: `tan 2x - cot x = sqrt3 - sqrt3 = 0`. OK
If `x = pi/2` --> `2x = pi` --> `tan 2x = tan pi = 0` --> `cot x = cot (pi/2) = 0`, then: `tan 2x - cot x = 0 - 0 = 0`. OK

Answer 2

Possible solutions in the interval `[0,2pi)` are `{pi/6,(5pi)/6,(7pi)/6,(11pi)/6}`

Explanation:

`tan2x-cotx=0`

`hArr(2tanx)/(1-tan^2x)=1/tanx`

or `2tan^2x=1-tan^2x`

or `3tan^2x-1=0`

i.e. `(sqrt3tanx+1)(sqrt3tanx-1)=0`

i.e. `tanx=1/sqrt3` or `-1/sqrt3`

Hence `x=npi+-pi/6`

and possible solutions in the interval `[0,2pi)` are

`{pi/6,(5pi)/6,(7pi)/6,(11pi)/6}`