To solve the equation 2sin(x) + 2cos(x) = 0, we can divide both sides of the equation by 2 to simplify it:

sin(x) + cos(x) = 0

Now, we can use the trigonometric identity sin(x) = cos(π/2 - x) to rewrite the equation:

cos(π/2 - x) + cos(x) = 0

Next, we can use the trigonometric identity cos(A) + cos(B) = 2cos((A + B)/2)cos((A - B)/2) to further simplify the equation:

2cos((π/2 - x + x)/2)cos((π/2 - x - x)/2) = 0

Simplifying this expression, we get:

2cos(π/4)cos(-x/2) = 0

Now, we have two possibilities for the equation to be true:

1) cos(π/4) = 0
2) cos(-x/2) = 0

For the first possibility, cos(π/4) = 1/√2, which is not equal to 0. Therefore, this possibility does not satisfy the equation.

For the second possibility, cos(-x/2) = 0, we can solve for x:

-x/2 = π/2 + nπ/2, where n is an integer

Multiplying both sides by -2, we get:

x = -π - 2nπ, where n is an integer

Therefore, the solution to the equation 2sin(x) + 2cos(x) = 0 is x = -π - 2nπ, where n is an integer.