Question

How do you solve `tanx-sqrt2=1/(tanx+sqrt2)` in the interval `0<=x<=2pi`?

Answer 1

`(tanx - sqrt(2))(tanx + sqrt(2)) = 1`

`tan^2x - 2 = 1`

`tan^2x= 3`

`tanx = +-sqrt(3)`

`x = pi/3, (2pi)/3, (4pi)/3 and (5pi)/3`

Hopefully this helps!