Question

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

Answer 1

`x in {pi/12, (5pi)/12, (7pi)/12,(11pi)/12,(13pi)/12,(17pi)/12,(19pi)/12,(23pi)/12}`

Explanation:

If `sec(4x) = 2`
then
`color(white)("XXX")4x=2pi*k+-pi/3` for `kinZZ`
`color(white)("XXXXXXXXXX")`this is based on the standard angles:
`color(white)("XXXXXXXXXX")cos(+-pi/3)=1/2`

`x=(pik)/2+-pi/12 = pi(6k+-1)/12`
`color(white)("XXX")`for all values of `k` which evaluate as `xin[0,2pi]`

`k=0 rarr x=pi/12` (other values give `x < 0`)

`k=1 rarr x= (5pi)/12 or (7pi)/12`

`k=2 rarr x=(11pi)/12 or (13pi)/12`

`k=3 rarr x=(17pi)/12 or (19pi)/12`

`k=4 rarr x=(23pi)/12` (other values give `x > 0`)