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How do you get the two sides in this inequality? 78 >= 80 - 10cos(pi(t)/12)?
78 >= 80 - 10cos(pi(t)/12)
10cos(pi(t)/12) >= 2
cos(pi(t)/12) >= 1/5
1.369 =< pi(t)/12 <= 4.914
5.23 =< t <= 18.77
I don't understand where the two sides in 1.369 =< pi(t)/12 <= 4.914 come from
1 Answer
- ?Lv 69 years agoFavorite Answer
Are there additional restrictions on t? For example, is 0 < t < 24, perhaps (thus limiting yourself to one period of the cosine)?
From cos(pi(t)/12) >= 1/5 , you should get two separate inequalities:
either (? 0 <= ) pi t/12 <= 1.369 or 4.914 <= pi t/12 ( <= 2*pi )
and so 0 <= t <= 5.23 or 18.77 <= t <= 24
(Your solution gives those t where 78 <= 80 - 10cos(pi(t)/12) instead.)
