Solve SIN(4X) - SIN(2x) = 0

Please help me solve all values for x within 0 and 2pi
Thanks!

Use theorem
Sin(2A) = 2·SinA·CosA


..... .... ... Sin(4x) - Sin(2x) = 0
.... ...... Sin(2*2x) - Sin(2x) = 0
2·Sin(2x)·Cos(2x) - Sin(2x) = 0 ..... ← Now, factor out Sin(2x)
... ... Sin(2x)*[2Cos(2x) - 1] = 0

This is true for:
i. Sin(2x) = 0 ..... ← Sine is 0 for angles 0,π,2π,3π,4π,5π,etc.
... So, 2x = 0,π,2π,3π,4π,5π,etc.
..... ..... x = 0, π/2, π, 3π/2, 2π

ii. 2Cos(2x) - 1 = 0
... ..... Cos(2x) = 1/2 ..... Cosine is 1/2 for angles π/3,5π/3,7π/3,11π/3,13π/3,etc.
..... ..... So, 2x = π/3,5π/3,7π/3,11π/3,13π/3,etc.
..... ...... ...... x = π/6, 5π/6, 7π/6, 11π/6

Therefore, from i. and ii.,
ANSWER
0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6, 2π

Have a good one!
.

sinA - sinB = 2sin (A-B)/2 cos(A+B)/2

hence ..

sin4X - sin2X = 2sin2Xcox6X = 0

=> sin(2X).cos(6X) = 0

hence .. case (i) sin(2X) = 0 .. which means X=0

case (ii) cos(6X) = 0 which means .. X = pi/2 or 90 deg

:)