Question

2cos^2x+sqrt(3)cosx=0 solution set: {pi/2, 3pi/2, 7pi/6, 5pi/6} I can't figure out how to get those solutions?

Answer 1

See the explanation below

Explanation:

The equation can be written as
`cos x * (2 * cos x + sqrt(3)) = 0`
which implies, either `cos x = 0 or 2 * cos x + sqrt(3) = 0`
If `cos x = 0` then the solutions are `x = pi /2 or 3*pi /2 or (pi / 2 + n * pi)`, where n is an integer
If `2 * cos x + sqrt(3) = 0, then cos x = -sqrt(3) / 2, x = 2 * pi / 3 +2 * n * pi or 4 * pi / 3 +2 * n * pi` where n is an integer

Answer 2

Solve `2cos^2 x + sqrt3.cos x = 0`

Explanation:

cos x(2cos x + sqrt3) = 0
a. cos x = 0 --> `x = pi/2` and `x = (3pi)/2` (Trig unit circle)
b. `cos x = - sqrt3/2` --> `x = +- (5pi)/6` (Trig unit circle)
Note. The arc `-(5pi)/6` is the same as the arc `(7pi)/6` (co-terminal)

Answers: `pi/2; (3pi)/2; (5pi)/6 and (7pi)/6`