Find all solutions to the equation in the interval [0, 2π).

sin 2x - sin 4x = 0

First we move sin4x to right side
sin(2x) = sin(4x) 
now we simplify 4x
sin(2x) = sin(2*2x) 
using double angle formula 
sin(2x) = 2sin(2x)cos(2x) 
subtracting sin2x on both sides 
2sin(2x)cos(2x) - sin(2x) = 0 
taking sin2x common
sin(2x) * (2cos(2x) - 1) = 0 
now using product rule 
so 
sin(2x) = 0
or
2cos(2x) - 1 = 0
cos(2x) = 1/2

hence 
x = 0 + nπ/2, π/6 + nπ, 5π/6 + nπ 

x = 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6, 2π

The correct answer would be "convenience"