To solve the equation 3sin^2x + 1 = 5sin2x + 2cos^2x = 1, we need to simplify the expression first.

Recall that sin^2(x) + cos^2(x) = 1 for all values of x.

Given the equation, we can write it as:

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

=> 3sin^2(x) + 1 = 10sin(x)cos(x) + 2 - 2sin^2(x)

=> 3sin^2(x) + 1 = 2 - 2sin^2(x) + 10sin(x)cos(x)

=> 5sin^2(x) + 10sin(x)cos(x) - 1 = 0

Now, we have a quadratic equation in terms of sin(x). Let's solve it using the quadratic formula:

sin(x) = [-b ± sqrt(b^2 - 4ac)] / 2a

Here, a = 5, b = 10, and c = -1.

sin(x) = [-10 ± sqrt(10^2 - 4 5 (-1))] / 2 * 5

sin(x) = [-10 ± sqrt(100 + 20)] / 10

sin(x) = [-10 ± sqrt(120)] / 10

sin(x) = [-10 ± 2√30] / 10

sin(x) = -1 ± √30 / 5

Thus, the solutions for sin(x) are sin(x) = (-1 - √30) / 5 and sin(x) = (-1 + √30) / 5.

Therefore, the solutions for the given equation are sin(x) = (-1 - √30) / 5 and sin(x) = (-1 + √30) / 5.