To prove this trigonometric identity, we can simplify both sides of the equation.

First, let's simplify the left side of the equation:

sin(π/2 + 3x)cos(2x) - 1
= sin(π/2)cos(2x)cos(3x) + cos(π/2)sin(2x)cos(3x) - 1
= 1(cos(2x)cos(3x)) + 0 - 1
= cos(2x)cos(3x) - 1

Now, let's simplify the right side of the equation:

sin(3x)cos(3π/2 - 2x)
= sin(3x)cos(π/2 + 2x)
= sin(3x)cos(π/2)cos(2x) - sin(3x)sin(π/2)sin(2x)
= sin(3x)(0)(cos(2x)) - sin(3x)(1)(sin(2x))
= -sin(3x)sin(2x)

Therefore, the left side (cos(2x)cos(3x) - 1) is not equal to the right side (-sin(3x)sin(2x)), and the trigonometric identity sin(π/2+3x)cos(2x)-1=sin(3x)cos(3π/2-2x) does not hold true.