that would be when
200+200cos[π/6(t+2)] = 0
cos[π/6(t+2)] = -1
π/6 (t+2) = kπ
t+2 = 6k
t = 6k-2
That is, for t = 4,10
Sales of snowmobiles are seasonal. Supposed sales in Camrose, Alberta are approximately by S=200+200cos[pi/6(t+2)], where t is time in months with t=0 corresponding to January. For what months are sales equal to 0.
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