To solve this equation, we need to first isolate the arccos function by dividing by 3 on both sides:

arccos(2x+3) = 5pi/6

Next, we need to find the angle whose cosine is equal to (2x+3):

cos(theta) = 2x+3

Since the cosine function has a range of [-1, 1], the value of 2x+3 must be between -1 and 1:

-1 <= 2x+3 <= 1
-4 <= 2x <= -2
-2 <= x <= -1

Now, we need to find the angle whose cosine is equal to (2x+3) using the inverse cosine function:

theta = arccos(2x+3) = 5pi/6

Using the property of inverse cosine, we get:

2x+3 = cos(5pi/6)
2x+3 = -1/2
2x = -1/2 - 3
2x = -7/2
x = -7/4

Therefore, the solution to the equation 3arccos(2x+3) = 5pi/2 is x = -7/4.