Question

How do you solve `sec^2x=4` in the interval `0<=x<=2pi`?

Answer 1

`pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

Explanation:

`sec^2 x = 1/(cos^2 x) = 4`
`cos^2 x = 1/4`
`cos x = +- 1/2`

a. `cos x = 1/2` --> `x = +- pi/3`
`x = (5pi)/3` is co-terminal to `x = -pi/3`
b. `cos x = -1/2` --> `x = +- (2pi)/3`
`x = (4pi)/3` is co-terminal to `x = - (2pi)/3`
Answers for `(0, 2pi):`
`pi/3, (2pi)/3, (4pi)/3, (5pi)/3`