Question

How to solve cosA+cos2A+cos3A=0 ?

Answer 1

`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`.