How to solve cosA+cos2A+cos3A=0 ?

1 Answer
Nghi N
Oct 20, 2017

#x = pi/4 + kpi#
#x = (3pi)/4 + kpi#
#x = (2pi)/3 + 2kpi#
#x = (4pi)/3 + 2kpi#

Explanation:

Use trig identity:
#cos x + cos y = 2 cos ((x+ y)/2)cos ((x - y)/2)#
In this case:
cos A + cos 3a = 2 cos 2a. cos a
Therefor:
cos A + cos 3A + cos 2A = 2cos (2A).cos A + cos (2A) =
= cos 2A(2cos A + 1) = 0.
Either factor should be zero:
a. cos 2A = 0 --> unit circle --> 2 solutions
#2A = pi/2 + 2kpi#, and #2A = (3pi)/2 + 2kpi#
1. #2A = pi/2 + 2kpi# --> #A = pi/4 + kpi#
2. #2A = (3pi)/2 + 2kpi #--> #A = (3pi)/4 + kpi#
b. (2cos A + 1) = 0 --> #cos A = - 1/2#
Trig table and unit circle give 2 solutions:
#A = +- (2pi)/3 + 2kpi#
#(-2pi)/3# is co-terminal to #(4pi)/3#.

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