To solve the equation 1 - cos(π+x) - sin(3π/2 + x/2) = 0, we need to simplify the trigonometric expressions and then solve for x.

1 - cos(π+x) - sin(3π/2 + x/2) = 0

Using the properties of trigonometric functions, we know that cos(π+x) = -cos(x) and sin(3π/2 + x/2) = -cos(π/2 - x/2).

Therefore, the equation becomes:

1 + cos(x) - cos(π/2 - x/2) = 0

Now, let's simplify cos(π/2 - x/2):

cos(π/2 - x/2) = cos(π/2)cos(x/2) + sin(π/2)sin(x/2)
cos(π/2) = 0 and sin(π/2) = 1, so:
cos(π/2 - x/2) = 0cos(x/2) + 1sin(x/2) = sin(x/2)

Now, substitute this back into the equation:

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

Rearranging the terms, we get:

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

Now, we need to use trigonometric identities to further simplify this expression. We can use the identity cos(x) = cos^2(x/2) - sin^2(x/2).

So the equation becomes:

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

Let y = sin(x/2), then cos(x/2) = √(1 - y^2).

Substitute these into the equation:

(√(1-y^2))^2 - y^2 - y = -1
1 - y^2 - y^2 - y = -1
-2y^2 - y + 2 = 0

This is a quadratic equation in y. Solving for y using the quadratic formula, we get two values for y:

y = (-(-1) ± √((-1)^2 - 4(-2)2)) / 2(-2)
y = (1 ± √(17)) / (-4)
y = (1 + √17)/(-4), (1 - √17)/(-4)

Now, substitute y back into y = sin(x/2) to find the corresponding values of sin(x/2). Then solve for x using these values.

Therefore, the solutions to the equation 1 - cos(π+x) - sin(3π/2 + x/2) = 0 are the values of x that make sin(x/2) equal to (1 + √17)/(-4) or (1 - √17)/(-4).