To solve the equation sin(2x + π/2) = sin(x - π/3), we can use the identity sin(a) = sin(b) if and only if a = b + 2πn or a = π - b + 2πn, where n is an integer.

First, let's simplify the equation by expanding the trigonometric functions:

sin(2x + π/2) = sin(x - π/3)

Using the identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b), we have:

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

Since cos(π/2) = 0 and sin(π/2) = 1, the equation becomes:

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

Simplifying further:

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

Now, let's use the double angle identities:

sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x)

Substituting these identities into the equation:

2sin(x)cos(x) + 1 - 2sin^2(x) = (1/2)sin(x) - (√3/2)cos(x)

Rearranging the terms:

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

Now, let's combine like terms:

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

Multiplying through by 2 to eliminate the fractions:

4sin(x)cos(x) - sin(x) + √3cos(x) - 4sin^2(x) + 2 = 0

Rearranging the terms:

4sin(x)cos(x) - sin(x) - 4sin^2(x) + √3cos(x) + 2 = 0

Now, let's factor out sin(x) and cos(x):

sin(x)(4cos(x) - 1) - 4sin^2(x) + √3cos(x) + 2 = 0

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

sin(x)(4cos(x) - 1) - 4(1 - cos^2(x)) + √3cos(x) + 2 = 0

Expanding and simplifying:

4sin(x)cos(x) - sin(x) - 4 + 4cos^2(x) + √3cos(x) + 2 = 0

Rearranging the terms:

4cos^2(x) + (4√3 + 1)cos(x) - 4sin(x) + 6 = 0

Now, we have a quadratic equation in terms of cos(x). We can solve this equation using the quadratic formula:

cos(x) = [-b ± √(b^2 - 4ac)] / (2a)

Plugging in the values:

cos(x) = [-(4√3 + 1) ± √((4√3 + 1)^2 - 4(4)(6))] / (2(4))

Simplifying further:

cos(x) = [-(4√3 + 1) ± √(48 - 96√3 + 1 - 96)] / 8

cos(x) = [-(4√3 + 1) ± √(-47 - 96√3)] / 8

Since the square root of a negative number is not real, there are no real solutions to the equation sin(2x + π/2) = sin(x - π/3).