Question

How do you solve the equation `sqrt3sectheta+2=0` for `-pi<=theta<=3pi`?

Answer 1

`(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3`

Explanation:

`sqrt3.sec t + 2 = 0`.
`sqrt3/(cos t) = - 2`
`sqrt3 = - 2cos t`
`cos t = - sqrt3/2`

Trig table and unit circle -->
a. For interval `(-pi, pi)` --> `cos t = -sqrt3/2` --> arc `t = +- (2pi)/3`
--> 2 solution arcs -->`t = (2pi)/3` and `t = (4pi)/3` (co-terminal)
b. For interval `(-pi, pi + 2pi)` or interval `(-pi, 3pi)`,
add 2 solution arcs -->
`t = (2pi)/3 + 2pi = (8pi)/3` and `t = (4pi)/3 + 2pi = (10pi)/3`

Answers for `(-pi, 3pi)`:
`(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3)`