An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house?

Answer: 5.2308694 ≤ t ≤ 18.766913”

F(t) = 80 - 10cos(pi*t/12), 0≤t≤24

78 ≤ 80 - 10cos(pi*t/12)

10cos(pi*t/12) ≤ 80 - 78

cos(pi*t/12) ≤ 0.2

There are two points on the Unit Circle that have cos(theta) = 0.2; they’re in Quadrants I and IV.

We want the CCW arc from the one in Quadrant I to the one in Quadrant IV because that’s where cos(theta) ≤ 0.2.

arccos(0.2) ≤ pi*t/12 ≤ 2 pi – arccos(0.2)

(12/pi)arccos(0.2) ≤ t ≤ (12/pi)(2 pi – arccos(0.2))

(12/pi)arccos(0.2) ≤ t ≤ 24 – (12/pi)arccos(0.2))

5.2308693978123 ≤ t ≤ 18.7691306021877

[You must have hit the “6” key twice by mistake, Sun.]

Here’s a GeoGebra check: http://www.wyzant.com/resources/files/267292/temperature_above_78

“I graphed F(t) in Y1 and 78 in Y2 on my graphing calculator to find the intersection points for the answer but I got 5.2308694 part and I can't get the 18.766913 part. Can someone please tell me how can I get the other part on my calculator?”

Check your viewing windows. Do they include x = 25?

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