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Since the equation involves trigonometric functions, the domain will generally be all real numbers. For simplicity, we can limit the domain to one period of each function and then find the general solution. The period of the function $$\sin^2(2x)$$ is the same as the period of the $$\sin(2x)$$ function since squaring doesn't affect the periodicity. Since the coefficient of x inside the sine function is 2, the period is given by $$\frac{2\pi}{2}=\pi$$. Similarly, the period of $$3\cos(2x) - 2$$ is the same as the period of the $$\cos(2x)$$ function. The coefficient of x inside the cosine function is also 2, making the period: $$\frac{2\pi}{2}=\pi$$. Therefore, we can limit the domain to $$[0, \pi]$$ to keep things simple. The range of both functions can be limited to $$[-1,1]$$ as sine and cosine functions range from $$-1$$ to $$1$$.

First, we need to graph the functions $$y = \sin^2(2x)$$ and $$y = 3\cos(2x) - 2$$ on the same set of axes. It would be helpful to use graphing software or a graphing calculator for this step. Remember that since we are dealing with trigonometric functions, we should keep the angle mode in radians. After plotting the graphs, we can see that there are two intersection points within the domain $$[0, \pi]$$.

Solve the equation at the intersection points: $$\sin^2(2x) = 3\cos(2x) - 2$$ We can use numerical methods or graphing tools to find the intersection point values of x within the domain $$[0, \pi]$$. The graphing calculator will give approximate values for the x-coordinates of the intersection points.

Once we find the approximate x-coordinates of the intersection points, we can write the general solution for the equation. Suppose that the x-coordinates of the intersection points are $$x_1$$ and $$x_2$$, then the general solution will be: $$x = x_1 + n\pi$$ $$x = x_2 + m\pi$$ where n and m are integers.