Find all solutions to cos(8x)-cos(2x)=sin(5x) on 0<x<π.

Answer 1

cos(8x) = cos(5x + 3x) = cos(5x)cos(3x) - sin(5x)sin(3x)

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

Then, cos(8x) - cos(2x) = -2sin(5x)sin(3x)

Then, -2sin(5x)sin(3x) = sin(5x) or

sin(5x)(1 + 2 sin(3x)) = 0

Thus, sin(5x) = 0 or sin(3x) = 1/2

As sin(x) = 0 for x = nπ

sin(5x) = 0 for x = nπ/5

For 0 < x < π, we have π/5, 2π/5, 3π/5, 4π/5

1 + 2 sin(3x) = 0 for sin(3x) = -1/2

As sin(x) = -1/2 for x = 7π/6 + 2nπ, 11π/6 + 2nπ

For n = 1, 7π/6 + 2nπ, 11π/6 + 2nπ or greater than 3π, so x/3 would be greater than π.

Thus, we have 7π/18, 11π/18

Our solution is π/5, 7π/18, 2π/5, 3π/5, 11π/18, 4π/5