Question

How do you solve `sec^2x-2=0`?

Answer 1

The general solution is `npi+-pi/3`

Explanation:

As `sec^2x-2=0`, we have

`sec^2x-(sqrt2)^2=0` or

`(secx+sqrt2)(secx-sqrt2)=0` and

`secx=+-sqrt2=+-sec(pi/3)`

Hence in the range `(-pi,+pi)` the values are `{+-pi/3,+-(2pi)/3}`, which can also be written as `{+-pi/3,+-(pi-pi/3)}`

Hence, the general solution is `npi+-pi/3`