To solve cos5x+cos3x+cosx = 0

Solution:

LHS: cos5x+cosx+cos 3x = 2cos (6x/2) sin(4x/2) +cos3x = 0. Or

2cos3x(sin2x+1) = 0. Or

cos3x = 0 Or 3x = 2npi. Or x = 2npi/3

sin2x +1 = 0 gives: sin2x =-1 . Or 2x = npi +(-1)^n*pi/3 Or

x = npi/2+(-1)^n*pi/3

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.**