Question

Sin2x-sin2xcos2x=sin4x

Answer 1

It looks like the given equation is

sin(2x) - sin(2x) cos(2x) = sin(4x)

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

which lets us rewrite the equation as

sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)

Move everything over to one side and factorize:

sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0

sin(2x) - 3 sin(2x) cos(2x) = 0

sin(2x) (1 - 3 cos(2x)) = 0

Then we have two families of solutions,

sin(2x) = 0   or   1 - 3 cos(2x) = 0

sin(2x) = 0   or   cos(2x) = 1/3

[2x = arcsin(0) + 2nπ   or   2x = π - arcsin(0) + 2nπ]

… … …   or   [2x = arccos(1/3) + 2nπ   or   2x = -arccos(1/3) + 2nπ]

(where n is any integer)

[2x = 2nπ   or   2x = π + 2nπ]

… … …   or   [2x = arccos(1/3) + 2nπ   or   2x = -arccos(1/3) + 2nπ]

[x = nπ   or   x = π/2 + nπ]

… … …   or   [x = 1/2 arccos(1/3) + nπ   or   x = -1/2 arccos(1/3) + nπ]