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.
To find the months when sales are equal to zero, we can equate the equation to zero and solve for t.
Given: S = 200 + 200cos[π/6(t+2)]
Setting S = 0:
0 = 200 + 200cos[π/6(t+2)]
Rearranging the equation:
200cos[π/6(t+2)] = -200
Dividing by 200:
cos[π/6(t+2)] = -1
To find the values of t for which cos[π/6(t+2)] = -1, we need to find the angles where cosine equals -1.
The cosine function equals -1 when the angle is (2n + 1)π, where n is an integer.
Solving for t:
π/6(t+2) = (2n + 1)π
Dividing by π/6:
t + 2 = (2n + 1)(6/π)
Subtracting 2:
t = (2n + 1)(6/π) - 2
Therefore, the sales are equal to zero in months when t is given by the equation:
t = (2n + 1)(6/π) - 2, where n is an integer.
To determine the months when sales are equal to 0, we need to solve the equation S = 0, where S represents the sales equation.
Given the equation: S = 200 + 200cos[pi/6(t+2)]
To find the months when sales are equal to 0, we set S equal to 0:
0 = 200 + 200cos[pi/6(t+2)]
Now we solve for t.
First, subtract 200 from both sides of the equation:
-200 = 200cos[pi/6(t+2)]
Next, divide both sides of the equation by 200:
-1 = cos[pi/6(t+2)]
Now, we need to take the inverse cosine (arccos) of both sides to isolate t+2:
arccos(-1) = arccos[cos[pi/6(t+2)]]
Since cos(x) is an even function, we can simplify:
pi = pi/6(t+2)
Divide both sides of the equation by pi/6:
6 = t + 2
Finally, isolate t by subtracting 2 from both sides:
t = 6 - 2
t = 4
Therefore, the sales in Camrose, Alberta are equal to 0 in the month of April (t = 4), based on the given equation.
when cos(pi/6 (t+2)) = -1
That is, where
pi/6 (t+2) = pi