Question

How do you solve `2cosx + sqrt3=0` for x in the interval [0,2pi)?

Answer 1

`(5pi)/6 and (7pi)/6`

Explanation:

`2cos x + sqrt3 = 0`
`cos x = -sqrt3/2`
Unit circle and trig table of special arcs give
`cos x = -sqrt3/2` --> `x = +- (5pi)/6`
`-(5pi)/6` --> co-terminal with `(7pi)/6`
Answers: `(5pi)/6 and (7pi)/6` (0 < x < `2pi`)