Question

How do you solve `sec^5x = 4secx` from 0 to 2pi?

Answer 1

Solve `sec^5 x = 4sec x`

Ans:`+- pi/4, +- (3pi)/4)`

Explanation:

`sec^5 x - 4sec x = 0`

`sec x(sec^4 x - 4) = sec x(sec^2 x - 2)(sec^2 x + 2) =`

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

a. `sec x = 1/cos x = 0` (undefined)
b. `1/cos x = sqrt2` --> `cos x = 1/sqrt2 = sqrt2/2` --> `x = +- pi/4`
c. `sec x = 1/cos x = -sqrt2` --> `cos x = -1/sqrt2` --> `x = +-(3pi)/4`
d. (sec^2 x + 2) (always positive)

Answers: `+- pi/4, +- (3pi)/4`
Note. `Arc -pi/4` is arc `(7pi)/4`, and arc `-(3pi)/4` is the arc `(5pi)/4`