We have to find the solution x for the equation cos x + cos 3x = 0.

Using the relation cos 3x = 4 ( cos x)^3 - cos x, we can write cos x + cos 3x = 0 as:

cos x + 4 ( cos x)^3 - cos x = 0

=> 4 ( cos x)^3 = 0

=> cos x = 0

=> x = arc cos 0

=> x = pi/2.

**As cos x is a periodic function the solutions include all values equal to pi/2 + 2k*pi, where k is an integer.**

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}.**