can someone please show and explain how to simplify (sin2x-sin6x)/(cos10x-cos6x)

recall that:
sinA + sinB = 2sin[(A + B)/2] cos[(A - B)/2]

0 ≤ x < 2π
then
0 ≤ 2x < 4π
0 ≤ 4x < 8π

sin(2x) + sin(4x) + sin(6x) = 0
let y = 2x
we have:
siny + sin2y + sin3y = 0
sin3y + siny + sin2y = 0
2sin[(3y + y)/2] cos[(3y - y)/2] + sin2y = 0
2sin2y cosy + sin2y = 0
sin2y (2cosy + 1) = 0
sin2y = 0, cosy = -1/2
sin4x = 0, cos2x = -1/2
4x = 0, π, 2π, 3π, 4π, 5π, 6π, 7π OR 2x = 2π/3, 4π/3, 8π/3, 10π/3
x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4 OR x = π/3, 2π/3, 4π/3, 5π/3
x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, π/3, 2π/3, 4π/3, 5π/3
x = 0, π/4, π/3, π/2, 2π/3, 3π/4, π, 5π/4, 4π/3, 3π/2, 7π/4, 5π/3 ⇦ ANSWER


ALTERNATE METHOD:
recall that:
sin²A + cos²A = 1
sin2A = 2sinAcosA
sin3A = -4sin³A + 3sinA

sin2x + sin4x + sin6x = 0 ==> siny + sin2y + sin3y = 0
siny + 2siny cosy - 4sin³y + 3siny = 0
2siny cosy - 4sin³y + 4siny = 0
siny cosy - 2sin³y + 2siny = 0
siny (cosy - 2sin²y + 2) = 0
siny (cosy - 2(1 - cos²y) + 2) = 0
siny (cosy - 2 + 2cos²y + 2) = 0
siny (cosy + 2cos²y) = 0
siny cosy (1 + 2cosy) = 0
2siny cosy (2cosy + 1) = 0
sin2y (2cosy + 1) = 0
sin2y = 0 OR cosy = -1/2
sin4x = 0 OR cos2x = -1/2