To solve the equation 6sin²(x-3π/2) - 3√2cosx = 0, we can use trigonometric identities to simplify it.

Recall the trigonometric identities:
sin^2(x) = 1 - cos^2(x)

Using this identity, we can rewrite the equation as:
6(1-cos^2(x-3π/2)) - 3√2cosx = 0

Distribute the 6:
6 - 6cos^2(x-3π/2) - 3√2cosx = 0

Rearrange the terms:
6 - 6cos^2(x-3π/2) = 3√2cosx

Divide by 3√2:
2√2 - 2cos^2(x-3π/2) = cosx

Now let y = x - 3π/2 for easier notation:
2√2 - 2cos^2(y) = cos(y)

Rewrite cos(y) in terms of sin(y):
2√2 - 2cos^2(y) = √(1 - sin^2(y))

Square both sides to get rid of the radical:
4(1 - sin^2(y)) = 8 - 8sin^2(y)

Expand and rearrange:
4 - 4sin^2(y) = 8 - 8sin^2(y)

Combine like terms:
4 = 4sin^2(y)

Divide by 4:
1 = sin^2(y)

Taking the square root of both sides:
sin(y) = ±1

This means that sin(y) = 1 or -1. To find the possible values of y, solve for y in each case:

Case 1: sin(y) = 1
This occurs when y = π/2 or 3π/2.

Case 2: sin(y) = -1
This does not have any solutions since the range of sin function is [-1, 1].

Now substitute back y = x - 3π/2:
x - 3π/2 = π/2 or x - 3π/2 = 3π/2

Solving for x in each case:
x = 2π or x = 3π

Therefore, the solution to the equation 6sin²(x-3π/2) - 3√2cosx = 0 is x = 2π or x = 3π.