Sin graph transformations & one application question

[Unkindled]

Hi there, if I could get some help with these two questions I'd really appreciate it.

1. Sketch y= -2sin2(x+π/6)+1 for x∈[-π,π]
2. The depth of water at the entrance to a harbour t hours after high tide is D metres, where D = p+q cos (rt)° for suitable constants p, q, r. At high tide the depth is 7m; at low tide, 6 hours later, the depth is 3m. Find how soon after low tide a ship that requires a depth of at least 4m of water will be able to enter the harbour.

For the 1st question I know how to sketch graphs with phase shift/vertical transformations but what confuses me is what they included in the solutions:


It doesn't make sense, why are they getting rid of -11π/6 and replacing it with 13π/6?
Also when they say 2.b, they're referring to this question's solution:


For the second question I've figured out the equation and it's D = 5+2cos(30t)° but I just don't know how to solve the bolded part.

As always, thanks for your time.

 

[willyengland]

The limits are x∈[-π,π]
-11/12 π - 2/12 π = -13/12 π, which is smaller than -π, so outside the limits.

2. Set D = 4 and resolve for t.

 

[Unkindled]

The limits are x∈[-π,π]
-11/12 π - 2/12 π = -13/12 π, which is smaller than -π, so outside the limits.

2. Set D = 4 and resolve for t.

Oh wait no sorry didn't mean to ask that, for the second question I was going to ask this instead of that:

The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a particular day is given by D(t)=10+3sin(πt/6), 0≤t≤24. Boats which need a depth of w metres are permitted to enter the harbour only if the depth of the water at the entrance is at least w metres for a continuous period of 1 hour. Find, correct to one decimal place, the largest value of w which satisfies this condition.

Also, I'm sorry but I still don't get the answer to the first one - I mean I get that it doesn't satisfy the criteria x∈[-π,π] but the numbers are just confusing. Could you maybe elaborate on it?

 

[willyengland]

Are you asking, why it is e.g. π/6?
Well, this is a value, you have to look up.
Or you know it by heart. π/6 = 30° and sin(30°) = 0.5

Then, due to the periodicity of the Sin-function, there are more values, where the Sinus is 0.5.
Those lie symmetrically around the maximum.
e.g. x₁ = π/6
Then x₂ = Max + |Max - x₁|
Max = π/2
which gives x₂ =π/2 + (π/2 - π/6) = 5π/6

The next maximum is at 2π + π/2 = 5π/2 = 15π/6
x₃ = 15π/6 - (π/2 - π/6) = 13π/6

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2. Set D = w and resolve for t.
This gives you the last possible time for exit: t₂
Since the function is symmetrical around the y-axis, the first possible time for entry is t₁ = -t₂
So the maximum time is 2t₂ minus 1 hour.