Question

How do you solve the equation equation is `tan^2x + secx = 1` for `0<x<2pi`?

Answer 1

The solutions are `pi,pi/3,(5pi)/3`

Explanation:

We have that

`tan^2x + secx = 1=>sec^2x-1+secx-1=0=>(secx+1)*(secx-2)=0`

Hence we have `secx=-1=>x=pi` and `secx=2=>x=pi/3 or x=(5pi)/3`

Remember that `0<=x<=2pi`