Question

3sin^2*2x + 2cos^2*2x-2cos*2x = 0

Answer 1

3sin^2(2x) + 2cos^2(2x)-2cos(2x) = 0
3(1-cos^2(2x)) + 2cos^2(2x)-2cos(2x) = 0
3-3cos^2(2x) + 2cos^2(2x)-2cos(2x) = 0
- cos^2(2x) – 2cos(2x) + 3 = 0
cos^2(2x) + 2 cos(2x) – 3 = 0
(cos(2x) + 3)*(cos(2x)  - 1) = 0

cos2x = 1
2x = 2pik
x = pik, k ∈Z