Find all exact solutions between [0,2pi). cos(8x)-cos(4x)=0

Answer 1

I just wanted to show a different approach:

If you folks learned your sum to product rules, you can get to your answers just as quickly.

In this case cos(x) - cos(y) = -2sin((x+y)/2)sin((x-y)/2)

(here's a link to one website that explains this if your interested: https://gateway.asurams.edu/affordable-learning-program/14-SumToProductAndProductToSumFormulas/14-Sum2ProductAndProduct2Sum.pdf)

Anyways,

cos(8x) - cos(4x) = -2sin((8x+4x)/2)sin((8x-4x)/2)

=-2sin(6x)sin(2x)

Thus, cos(8x) - cos(4x) = 0

becomes: -2sin(6x)sin(2x) = 0

and by zero product property, we have

sin(6x) = 0 or sin(2x) = 0

which is solved practically the same way as shown in the previous response. You'll end up with the same answers.

Just thought I'd add that in.

Mr. K

Answer 2

Recall: cos(2θ) = 2cos²θ - 1

So, cos(8θ) - cos(4θ) = 2cos²(4θ) - 1 - cos(4θ) = 0

2cos²(4θ) - cos(4θ) - 1 = 0

(2cos(4θ) +1)(cos(4θ) - 1) = 0

cos(4θ) = 1 or cos(4θ) = -1/2

If cos(4θ) = 1, then 4θ = 2kπ . So, θ = kπ/2 (k = 0, ±1, ±2, ...)

Solutions in [0, 2π): θ = 0, π/2, π, 3π/2

If cos(4θ) = -1/2, then 4θ = 2π/3 + 2kπ or 4π/3 + 2kπ

So, θ = π/6 + kπ/2 or π/3 + kπ/2

Solutions in [0, 2π): π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6

Solution set of original equation:

{ 0, π/2, π, 3π/2, π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6}