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
Srinivas
Oct 23, 2015

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

Nghi N.
Oct 23, 2015

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#

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