# How to solve cosA+cos2A+cos3A=0 ?

Oct 20, 2017

$x = \frac{\pi}{4} + k \pi$
$x = \frac{3 \pi}{4} + k \pi$
$x = \frac{2 \pi}{3} + 2 k \pi$
$x = \frac{4 \pi}{3} + 2 k \pi$

#### Explanation:

Use trig identity:
$\cos x + \cos y = 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$
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
$2 A = \frac{\pi}{2} + 2 k \pi$, and $2 A = \frac{3 \pi}{2} + 2 k \pi$
1. $2 A = \frac{\pi}{2} + 2 k \pi$ --> $A = \frac{\pi}{4} + k \pi$
2. $2 A = \frac{3 \pi}{2} + 2 k \pi$--> $A = \frac{3 \pi}{4} + k \pi$
b. (2cos A + 1) = 0 --> $\cos A = - \frac{1}{2}$
Trig table and unit circle give 2 solutions:
$A = \pm \frac{2 \pi}{3} + 2 k \pi$
$\frac{- 2 \pi}{3}$ is co-terminal to $\frac{4 \pi}{3}$.