How to solve cosA+cos2A+cos3A=0 ?

1 Answer
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.