Question

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

sin 2x - sin 4x = 0

Answer 1

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