To solve the equation 6 sin x cos x - 8 cos 2x = 0, we can use trigonometric identities to simplify it.

First, note that cos 2x = 2(cos x)^2 - 1. We can substitute this into the equation:

6 sin x cos x - 8(2(cos x)^2 - 1) = 0
6 sin x cos x - 16(cos x)^2 + 8 = 0

Now, we can use the double angle formula sin 2x = 2sin x cos x to simplify it further:

6(2sin x cos x) - 16(cos x)^2 + 8 = 0
12sin x cos x - 16(cos x)^2 + 8 = 0

Now, let's set cos x = u to make substitution:

12sin x u - 16u^2 + 8 = 0
12sin x u = 16u^2 - 8
3sin x u = 4u^2 - 2

Divide this equation by u:

3sin x = 4u - 2/u

Using the identity sin x = sqrt(1 - cos^2 x), we have:

3(sqrt(1 - u^2)) = 4u - 2/u

Squaring both sides:

9(1 - u^2) = 16u^2 - 8 + 4
9 - 9u^2 = 16u^2 - 4
25u^2 = 13
u = ±√(13/25)
u = ±√13/5

Therefore, the solutions for cos x are ±√13/5.