To solve the equation cos^2(3x) + 3sin(3x) = 3, we can use the trigonometric identity cos^2(x) + sin^2(x) = 1.

First, we rewrite the equation using this identity:

cos^2(3x) + 3sin(3x) = cos^2(3x) + sin(3x) + sin(3x) = cos^2(3x) + 2sin(3x) = 3

Next, we can rewrite cos^2(3x) in terms of sin(3x) using the identity cos^2(x) + sin^2(x) = 1:

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

Now, let's substitute y = sin(3x) to simplify the equation:

1 - y^2 + 2y = 3

Rearranging terms gives us:

y^2 - 2y - 2 = 0

Now we need to solve this quadratic equation for y. We can use the quadratic formula:

y = [ 2 ± sqrt(2^2 - 41(-2)) ] / 2*1
y = [ 2 ± sqrt(4 + 8) ] / 2
y = [ 2 ± sqrt(12) ] / 2
y = [2 ± 2√3] / 2
y = 1 ± √3

Therefore, sin(3x) = 1 ± √3

To find the values of x, we need to determine the possible values of sin(3x) given that it equals 1 ± √3. Since the range of sin function is -1 to 1, we can see that sin(3x) can only be 1 + √3.

So, sin(3x) = 1 + √3

From here, we can solve for x by taking the inverse sine of both sides:

3x = sin^(-1)(1 + √3)

x = sin^(-1)(1 + √3) / 3

Therefore, x = sin^(-1)(1 + √3) / 3.