To solve for x in the equation 2sin^3(x) - 3sin^2(x) + 5sin(x) = 0, we can first factor out sin(x) from the equation:

sin(x) * (2sin^2(x) - 3sin(x) + 5) = 0

Now we have two separate factors:

1) sin(x) = 0
2) 2sin^2(x) - 3sin(x) + 5 = 0

For the first factor, sin(x) = 0 at x = 0, π.

Now let's solve the quadratic equation in the second factor:

2sin^2(x) - 3sin(x) + 5 = 0

We can solve this using the quadratic formula:

sin(x) = [3 ± √(3^2 - 425)]/(2*2)
sin(x) = [3 ± √(9 - 40)]/4
sin(x) = [3 ± √(-31)]/4

Since the square root of a negative number is not a real number, sin(x) does not have real solutions for the second factor. Therefore, the solutions to the equation 2sin^3(x) - 3sin^2(x) + 5sin(x) = 0 are x = 0 and x = π.