To solve the equation 4cos(x/2) + cosx + 1 = 0, we can first make a substitution to simplify the equation. Let's replace cos(x/2) with u:

u = cos(x/2)

Now the equation becomes:

4u + 2u^2 + 1 = 0

Rearranging the terms, we get a quadratic equation in terms of u:

2u^2 + 4u + 1 = 0

We can solve this quadratic equation using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / 2a

Where a = 2, b = 4, and c = 1. Plugging these values in, we get:

u = (-4 ± √(4^2 - 421)) / 2*2
u = (-4 ± √(16 - 8)) / 4
u = (-4 ± √8) / 4
u = (-4 ± 2√2) / 4

Now we have two possible solutions for u:

u1 = (-4 + 2√2) / 4 = (2 - √2) / 2u2 = (-4 - 2√2) / 4 = (2 + √2) / 2

Now we need to find x, since we made the substitution u = cos(x/2). We can solve for x by using the relationship between u and cos(x/2):

u = cos(x/2)
cos(x/2) = (2 - √2) / 2

To find the values of x, we need to consider both u1 and u2:

When u = (2 - √2) / 2:
cos(x/2) = (2 - √2) / 2
x/2 = arccos((2 - √2) / 2) or x/2 = -arccos((2 - √2) / 2)

When u = (2 + √2) / 2:
cos(x/2) = (2 + √2) / 2
x/2 = arccos((2 + √2) / 2) or x/2 = -arccos((2 + √2) / 2)

These are the possible values of x that satisfy the original equation 4cos(x/2) + cosx + 1 = 0.