Can you help me with problems "d" and "e" of the following trig app question?

Question

anonymous · Answer

@Callisto

callisto · Answer

For (d), solve sin[0.2094(t-30)]=sin 270. Hmm.. I'm not sure

callisto · Answer

Since, sin 270 is the minimum for a normal sine curve. So, equate the sine part in the equaton and sin270. I think you'll get something

anonymous · Answer

Does this work?

Set
3π/2=(.2094)t
t=22.5 minutes or 22.5+(n)(30) where n is a positive integer.

callisto · Answer

[0.2094(t-30)]= 3pi /2
You have a horizontal shift

anonymous · Answer

Does that mean its wrong?

callisto · Answer

I think.. [0.2094(t-30)]= 3pi /2 would be more appropriate.. Not sure

anonymous · Answer

Wait so the answer is 3pi/2?

callisto · Answer

Nope, that is the equation only you need to solve t

anonymous · Answer

I dont uderstand though, [0.2094(t-30)] equals to 3pi /2. What I am to solve then?

callisto · Answer

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