Verify the identity:

cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

By using trigonometric identities, we can solve the identity given above. We just need to work on one side and make it equal to the other side. We choose the right side:

2−2sin2(2x)−2sin2(x)=1+1−2sin2(2x)−2sin2(x) = (1−2sin2(2x))+(1−2sin2(x)) + (1−2sin2(2x))+(1−2sin2(x)) =cos(4x)+cos(2x)

cos 4x + cos 2x = cos 4x + cos 2x

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.

It’s hard to understand what you put.. can you show a picture ?