Please Help me with this problem.

Question

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2theta). (Enter your answers as a comma-separated list.)

sin(3theta) - sin(6theta) = 0

Answer 1

sin(3x) - sin(6x) = 0; sin(3x) - 2sin(3x)cos(3x) = 0, sin(3x) = 0, 3x = π n, x = πn/3. So if n = 0, 1, 2, 3, 4, 5

We have 6 solutions: 0, π/3, 2π/3, π, 4π/3 and 5π/3 on interval [0, 2π).

Then 1 - 2cos(3x) = 0; cos(3x) = 1/2; 3x = π/3 + 2πn or 3x = 5π/3 + 2πn

x = π/9 + 2πn/3 and for n = 0, 1, 2 we have 3 solutions: π/9, 7π/9, 13π/9 on interval [0. 2π)

x = 5π/9 + 2πn/3 and for n = 0, 1, 2 we have 3 solutions: 5π/9, 11π/9, 17π/9 on interval [0,2π)

Total we have 12 solutions.

Answer 2

sin 3θ - sin [2(3θ)] = 0

sin 3θ - 2 sin 3θ cos 3θ = 0

sin 3θ = 2 sin 3θ cos 3θ

1 = 2 cos 3θ

Can you finish and answer?