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.
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
To find the months when sales are equal to 0, we need to solve the given equation:
S = 200 + 200cos[(π/6)(t+2)]
Where S represents the sales.
Setting S equal to 0, we have:
0 = 200 + 200cos[(π/6)(t+2)]
Next, let's isolate the cosine term by subtracting 200 from both sides of the equation:
-200 = 200cos[(π/6)(t+2)]
Dividing both sides by 200:
-1 = cos[(π/6)(t+2)]
To find the values of t, we need to find the values of (π/6)(t+2) where the cosine function equals -1.
Now, the cosine function equals -1 at π radians:
π = (π/6)(t+2)
Let's solve for t:
t+2 = 6
t = 4
Therefore, we have found that for t = 4, sales are equal to 0.
Since t = 0 corresponds to January, t = 4 corresponds to May.
Hence, sales are equal to 0 in the month of May.