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 <= -4 <= 2x <= --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/62x+3 = -1/2x = -1/2 - 2x = -7/x = -7/4
Therefore, the solution to the equation 3arccos(2x+3) = 5pi/2 is x = -7/4.
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 <=
-4 <= 2x <= -
-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/
2x = -1/2 -
2x = -7/
x = -7/4
Therefore, the solution to the equation 3arccos(2x+3) = 5pi/2 is x = -7/4.