To solve the equation 2sin^2x - 1 + cos3x = 0, we can use trigonometric identities to simplify it.

First, let's rewrite cos3x using the identity cos3x = 4cos^3x - 3cosx:

2sin^2x - 1 + 4cos^3x - 3cosx = 0

Rearranging the terms, we have:

2sin^2x + 4cos^3x - 3cosx - 1 = 0

Now, let's use the identity sin^2x + cos^2x = 1 to replace sin^2x in the equation:

2(1 - cos^2x) + 4cos^3x - 3cosx - 1 = 0

Expanding and rearranging the terms, we get:

2 - 2cos^2x + 4cos^3x - 3cosx - 1 = 0

Combining like terms, we have:

-2cos^2x + 4cos^3x - 3cosx + 1 = 0

Now, let's factor out a common factor of cosx:

cosx(-2cosx + 4cos^2x - 3) + 1 = 0

Now, we have two possibilities:

1) cosx = 0:
If cosx = 0, then x = π/2 + nπ, where n is an integer.

2) -2cosx + 4cos^2x - 3 = 0:
Rearranging the terms, we get:
4cos^2x - 2cosx - 3 = 0

Now, we can solve this quadratic equation for cosx. Factoring, we have:

(2cosx + 1)(2cosx - 3) = 0

Setting each factor equal to zero, we get:

2cosx + 1 = 0 or 2cosx - 3 = 0

Solving these equations, we have:

2cosx = -1 or 2cosx = 3

For 2cosx = -1, we have cosx = -1/2, which gives us x = 2π/3 + 2nπ or x = 4π/3 + 2nπ, where n is an integer.

For 2cosx = 3, we have cosx = 3/2, which has no real solutions since the range of cosine is -1 to 1.

Therefore, the solutions to the equation 2sin^2x - 1 + cos3x = 0 are:
x = π/2 + nπ, 2π/3 + 2nπ, or 4π/3 + 2nπ, where n is an integer.