Solve the equation cos5x+cos3x+cosx=0.

We'll notice the fact that we have a sum of 3 cosine
functions, so we can transform the sum of 2 of them into a
product.

We'll group the first term with the last and we'll
transform their sum into a product:

cos 5x + cos x = 2cos
[(5x+x)/2]cos[(5x-x)/2]

cos 5x + cos x =
2cos(6x/2)cos(4x/2)

cos 5x + cos x = 2 cos3x
cos2x

cos5x+cos3x+cosx = 2 cos3x cos2x +
cos3x

We'll notice the common factor
cos3x:

cos3x(2cos2x + 1) =
0

cos3x
=cos(2x+x)=cos2xcosx-sin2xsinx

cos2xcosx-sin2xsinx =
(2(cosx)^2 -1)cosx - 2cosx(sinx)^2

We'll transform (sinx)^2
= 1-(cosx)^2 and we'll open the
brackets:

2(cosx)^3-cosx-2cosx+2(cosx)^3=4(cosx)^3-3cosx

So,
cos3x(2cos2x + 1) = 0 will be written
as:

(4(cosx)^3-3cosx){2[2(cosx)^2 -1] +
1}=0

cosx(4(cosx)^2-3)(4(cosx)^2-1)=0

cos
x = 0, so x=pi/2 and
3pi/2

4(cosx)^2-3=0

4(cosx)^2=3

(cosx)^2=3/4

cosx=+/-sqrt3/2

x=pi/6,
11pi/6 and x=5pi/6,
7pi/6

4(cosx)^2-1=0

4(cosx)^2=1

cosx=+/-1/2

x=pi/3,5pi/3
and x=2pi/3, 4pi/3.