We are given `y=1747.29+1636.95sin(.52x-1.71)`

(1) You are correct about the period. The period will be `p=(2pi)/.52=(50pi)/13~~12.083` This is what it is -- we might like to have "nice" answers but that rarely happens in the real world. The problem is what value to approximate to.

(2) If y=2011.32:

`2011.32=1747.29+1636.95sin(.52x-1.71)`

`264.03=1636.95sin(.52x-1.71)`

`.1612938697=sin(.52x-1.71)`

(a) `"Sin"^(-1)(sin(.52x-1.71))="Sin"^(-1)(.1612938697)`

`.52x-1.71=.1620015487`

`.52x=1.872001549`

`x=3.600002978` thus your answer of x=3.6

(b) When using the arcsin (inverse sin) there is the possibility that the angle is in another quadrant, so you also have:

`"Sin"^(-1)(sin(.52x-1.71))=pi-"Sin"^(-1)(.1612938697)`

`.52x-1.71=2.979591105`

`.52x=4.689591105`

`x=9.018444432`

Thus to a good approximation we have two points where the y value is 2011.32: x=3.6 and x=9.02.

To find more, we add the period to the two possible answers:

`y=2011.32==>x~~3.6+-12.083` or `x~~9.018+-12.083`

The graph: