Question

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

Answer 1

@Callisto

Answer 2

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

Answer 3

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

Answer 4

Does this work?

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

Answer 5

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

Answer 6

Does that mean its wrong?

Answer 7

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

Answer 8

Wait so the answer is 3pi/2?

Answer 9

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

Answer 10

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