To solve the equation sin(x) + cos(2x) = 0, we can use some trigonometric identities.

First, let's rewrite cos(2x) using the double angle formula:
cos(2x) = 1 - 2sin^2(x)

Now, substitute this back into the equation:
sin(x) + 1 - 2sin^2(x) = 0

Rearrange the terms:
-2sin^2(x) + sin(x) + 1 = 0

Multiply the equation by -1 to make the leading coefficient positive:
2sin^2(x) - sin(x) - 1 = 0

Now, we have a quadratic equation in terms of sin(x). We can solve this by factoring or using the quadratic formula.

Factoring:
(2sin(x) + 1)(sin(x) - 1) = 0

Setting each factor equal to zero:
2sin(x) + 1 = 0 or sin(x) - 1 = 0

For 2sin(x) + 1 = 0:
2sin(x) = -1
sin(x) = -1/2

For sin(x) - 1 = 0:
sin(x) = 1

Now, we need to find the values of x that satisfy these equations.

For sin(x) = -1/2:
x = 7π/6 + 2πn or x = 11π/6 + 2πn (where n is an integer)

For sin(x) = 1:
x = π/2 + 2πn (where n is an integer)

Therefore, the solutions to the equation sin(x) + cos(2x) = 0 are:
x = 7π/6 + 2πn, 11π/6 + 2πn, π/2 + 2πn (where n is an integer)