To solve this equation, let's rewrite it in terms of sin(2x):

Let sin(2x) = y

The equation becomes:

3y^2 + 10y + 3 = 0

This is a quadratic equation in terms of y. To solve for y, we can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

Where a = 3, b = 10, and c = 3.

Plugging in these values, we get:

y = (-10 ± √(10^2 - 4 3 3)) / 2 * 3
y = (-10 ± √(100 - 36)) / 6
y = (-10 ± √64) / 6
y = (-10 ± 8) / 6

This gives us two possible solutions for y:

y = (-10 + 8) / 6 = -2 / 6 = -1/3
y = (-10 - 8) / 6 = -18 / 6 = -3

Since sin(2x) cannot be outside the range of [-1, 1], the solution y = -3 is not valid.

Thus, sin(2x) = -1/3.

Now, we need to find the values of x that satisfy this:

sin(arcsin(-1/3)) = -1/3

Thus, the solution is:

2x = arcsin(-1/3) + 2πn
x = (arcsin(-1/3) + 2πn) / 2

Where n is an integer.