Question

How do you solve `9+cot^2x=12` for `0<=x<=2pi`?

Answer 1

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

Explanation:

`9+cot^2x=12`

`cot^2x=12-9`

`cot^2x =3`

`1/tan^2x =3`

`1=3tanx`

`1/3 = tan^2x`

`tanx = +- 1/sqrt(3)`

`x = pi/6, (5pi)/6, (7pi)/6 and (11pi)/6`

Hopefully this helps!