Circular functions word problem

I was able to work out a, b, and c, but got stuck at d and c.i and c.ii
At a certain time of the year the depth of water d meters in the harbour at Bunk Island is given by the rule d = 3 + 1.8 cos(pi/6t), where t is the time in hours after 3 am.
a. Sketch the graph of the function d = 3 + 1.8 cos(pi/6t) over a 24-our period from 3 am to 3am.
b. At what time(s) does the high tide occur for t is a set of [0,24]?
c. At what time(s) does low tide occur for t is a set of [0,24]?

A passenger ferry operates between Main Beach and Bunk Island. It takes 50 minutes to go from Main Beach to Bunk Island. The ferry only runs between the hours of 8 am and 8 pm and is only able to enter the harbour at Bunk Island if the depth of water is at least 2 meters.

d. What is the earliest time the ferry should leave Main Beach so that it arrives at Bunk Island and can imemdiately enter the harbour?
Answer: 10:03am

e. The time to go from Bunk Island to Main Beach is also 50 minutes. The minimum time the ferry takes at Bunk Island harbor is 5 minutes. The minimum time at Main Beach is also 5 minutes.
i. What is the latest time the ferry can leave Main Beach to complete a round trip in 105 minutes?
ANSWER: 6:12pm

ii. How many complete round trips can the ferry make in a day?
ANSWER: 5 trips

b. High tide at t = 0, t = 12 (3 a.m. and 3 p.m.)
c. Low tide at t = 6, t = 18 (9 a.m. and 9 p.m.)

To find solutions for d and e, we need to find when d ≥ 2

3 + 1.8 cos(π/6 t) = 2
1.8 cos(π/6 t) = -1
cos(π/6 t) = -1/1.8 = -5/9

π/6 t = 2πk - arccos(-5/9), 2πk + arccos(-5/9)
t = 12k - 6/π arccos(-5/9), 12k + 6/π arccos(-5/9)

For t on interval [0, 24], we get:
t = 0 + 6/π arccos(-5/9) = 4.12497
t = 12 - 6/π arccos(-5/9) = 7.87503
t = 12 + 6/π arccos(-5/9) = 16.12497
t = 24 - 6/π arccos(-5/9) = 19.87503

Since we have high tide at t = 0 and t = 12 and low tide at t = 6 and 18: