To simplify this expression, we can use the trigonometric identity cos²θ + sin²θ = 1.

Given expression: 2cos²(π/2+x) + √3 sinx = 0

Using the identity cos(π/2 + x) = -sin(x), we can rewrite the expression as follows:

2(-sinx)² + √3 sinx = 0
2sin²x + √3 sinx = 0

Now, let's factor out sinx:

sinx(2sinx + √3) = 0

Setting each factor equal to zero:

sinx = 0 or 2sinx + √3 = 0

If sinx = 0, then x = arcsin(0) = nπ, where n is an integer.

If 2sinx + √3 = 0:
2sinx = -√3
sinx = -√3/2

This occurs when x = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.

Therefore, the solutions are:
x = nπ, where n is an integer, or
x = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.