To find when the population size will be 4800 during the first 7 years, we need to solve the equation given by:
[tex]\[ p(t) = 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) \][/tex]
We set this equal to 4800 to find the time \( t \):
[tex]\[ 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 \][/tex]
First, solve for \( \cos\left(\frac{2 \pi}{7} t\right) \):
[tex]\[ 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 \][/tex]
Subtract 4249 from both sides of the equation:
[tex]\[ - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 - 4249 \][/tex]
Simplify the right side:
[tex]\[ - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 551 \][/tex]
Now, divide both sides by -1180 to isolate the cosine term:
[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = \frac{551}{-1180} \][/tex]
[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = -\frac{551}{1180} \][/tex]
This further simplifies to:
[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = -0.4678... \][/tex]
Next, find the inverse cosine of -0.4678 to solve for \( \frac{2 \pi}{7} t \):
[tex]\[ \frac{2 \pi}{7} t = \cos^{-1}(-0.4678) \][/tex]
Using a calculator to find \( \cos^{-1}(-0.4678) \):
[tex]\[ \frac{2 \pi}{7} t \approx 2.2913 \text{ radians} \][/tex]
Now, solve for \( t \):
First instance:
[tex]\[ t = \frac{7}{2 \pi} \cdot 2.2913 \][/tex]
Calculate \( t \):
[tex]\[ t \approx 2.29 \text{ years} \][/tex]
There is another solution because cosine is periodic with period \( 2 \pi \). The other solution within one period (0 to 7 years) would be found as follows:
[tex]\[ \frac{2 \pi}{7} t = 2 \pi - 2.2913 \][/tex]
[tex]\[ \frac{2 \pi}{7} t \approx 3.9919 \text{ radians} \][/tex]
Solve for \( t \):
[tex]\[ t = \frac{7}{2 \pi} \cdot 3.9919 \][/tex]
Calculate \( t \):
[tex]\[ t \approx 4.71 \text{ years} \][/tex]
So, the solutions to the given problem are \( t = 2.29 \) years and \( t = 4.71 \) years.
Thus, the population will be 4800 at:
[tex]\[ t = 2.29 \text{ years} \][/tex]
or
[tex]\[ t = 4.71 \text{ years} \][/tex]
