`cot 2theta= 1/(tan 2 theta)`, so we have:

`tan 2 theta-1/(tan 2 theta)=0`

`tan^2 2θ = 1`

`tan 2θ = ± 1`

 

Since `0 ≤ θ < 2π`, we need to consider values of `2theta` such that `0 ≤ 2θ < 4π`. Hence, solving the above equation, we have:

`2 theta=pi/4,(3pi)/4,(5pi)/4(7pi)/4,` `(9pi)/4,(11pi)/4,` `(13pi)/4,(15pi)/4`

Dividing throughout by 2 gives us the full set of solutions in the required domain, `0 <= theta <2pi`:

`theta=pi/8,(3pi)/8,` `(5pi)/8,(7pi)/8,` `(9pi)/8,(11pi)/8,` `(13pi)/8,(15pi)/8`