solve: - cos 2x + cos x + cos 5x = 0

Question

I tried to use trig identities, first the double angle formula to subside -cos 2x

1. -2 cos^2 - 1 + cos x + cos 5x = 0
2. however I am lost on what to do next, should I have done something different?
3. any help would be obliged, thank you.

Answer 1

isn;t

cos(x) + cos(5x)=2cos(3x)cos(2x)

then problem becomes

cos(2x){2cos(3x)-1]=0

x₁=(π/4)+πk₁

x₂=(-π/4)+πk₂

k₁,k₂=0,+1,-1,...

3x₃=(π/3)+2πk,,,,x₃=(π/9)+2πk₃/3

3x₄=(-π/3)+2πk,,,,x₄=(-π/9)+2πk₄/3

k₃,k₄=0,+1,-1,...

Answer 2

Apply the formula

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

to the sum cos(x) + cos(5x)

This gives you new equation

-cos(2x) + 2cos(3x)cos(2x) = 0

cos(2x)(2cos(3x) - 1) = 0

This product is 0 iff at least one of the factors is 0.

Case 1:

cos(2x) = 0

2x = π/2 + kπ for any integer k

x = π/4 + kπ/2 for any integer k

Case 2:

2cos(3x) - 1 = 0

cos(3x) = 1/2

3x = π/3 + 2kπ for any integer k

or

3x = -π/3 + 2kπ for any integer k

So in conclusion there are an infinity of solutions falling into three form:

x = π/4 + kπ/2 for any integer k

x = π/9 + 2kπ/3 for any integer k

x = -π/9 + 2kπ/3 for any integer k