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?

2 Answers
Oct 23, 2015

See the explanation below

Explanation:

The equation can be written as
cos x * (2 * cos x + sqrt(3)) = 0cosx(2cosx+3)=0
which implies, either cos x = 0 or 2 * cos x + sqrt(3) = 0cosx=0or2cosx+3=0
If cos x = 0 cosx=0 then the solutions are x = pi /2 or 3*pi /2 or (pi / 2 + n * pi)x=π2or3π2or(π2+nπ), 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 * pi2cosx+3=0,thencosx=32,x=2π3+2nπor4π3+2nπ where n is an integer

Oct 23, 2015

Solve 2cos^2 x + sqrt3.cos x = 02cos2x+3.cosx=0

Explanation:

cos x(2cos x + sqrt3) = 0
a. cos x = 0 --> x = pi/2x=π2 and x = (3pi)/2x=3π2 (Trig unit circle)
b. cos x = - sqrt3/2cosx=32 --> x = +- (5pi)/6x=±5π6 (Trig unit circle)
Note. The arc -(5pi)/65π6 is the same as the arc (7pi)/67π6 (co-terminal)

Answers: pi/2; (3pi)/2; (5pi)/6 and (7pi)/6π2;3π2;5π6and7π6