What are the solutions of the equation: cos x + cos 3x = 0?

We'll transform the sum in product using the
formula:

cos a + cos b = 2 cos[(a+b)/2]cos
[(a-b)/2]

We'll write the formula for a = x and b =
3x:

cos x + cos 3x = 2cos[(x+3x)/2]cos
[(x-3x)/2]

cos x + cos 3x = 2cos[(4x)/2]cos
[(-2x)/2]

cos x + cos 3x = 2cos[(2x)]cos
[(-x)]

Since the function cosine is even, we'll write cos
[(-x)] = cos x.

cos x + cos 3x = 2cos 2x*cos
x

We'll put 2cos 2x*cos x =
0

cos 2x = 0

2x = +/- arccos 0
+ 2kpi

2x = +/-pi/2 +
2kpi

x = +/- pi/4 +
kpi

cos x =
0

x = +/- pi/2 +
2kpi

The solutions of the
equation are: {+/- pi/4 + kpi}U{+/- pi/2 + 2kpi}.