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

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

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Answer 1 · 2016-02-15T05:32:01

#(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#)

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

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