The pressure of a fluid flowing through a closed system, \( P \), after \( t \) seconds can be modeled by the function \( P(t)=100+40 \sin \left(\frac{9 \pi}{4} t\right) \). Answer parts (a) through (c) below. (a) In the interval \( [0,1] \), determine the times at which the pressure of the fluid is 100 mmHg . The solution set is \( \{\square \). (Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)

To find the times \( t \) at which the pressure \( P(t) = 100 \) mmHg in the interval \( [0, 1] \), we set the equation: \[ 100 + 40 \sin\left(\frac{9 \pi}{4} t\right) = 100 \] Subtracting 100 from both sides gives us: \[ 40 \sin\left(\frac{9 \pi}{4} t\right) = 0 \] Dividing by 40 yields: \[ \sin\left(\frac{9 \pi}{4} t\right) = 0 \] The sine function equals zero at integer multiples of \( \pi \): \[ \frac{9 \pi}{4} t = n\pi \quad \text{for integers } n \] Solving for \( t \): \[ t = \frac{4n}{9} \] Now we substitute integer values to find \( t \) within the interval \( [0, 1] \): - For \( n = 0 \): \[ t = \frac{4 \cdot 0}{9} = 0 \] - For \( n = 1 \): \[ t = \frac{4 \cdot 1}{9} \approx 0.44 \] - For \( n = 2 \): \[ t = \frac{4 \cdot 2}{9} \approx 0.89 \] - For \( n = 3 \): \[ t = \frac{4 \cdot 3}{9} \approx 1.33 \quad \text{(This is outside the interval [0,1])} \] Therefore, the times at which the pressure of the fluid is 100 mmHg in the interval \( [0, 1] \) are: \[ \{0, 0.44, 0.89\} \] The solution set is \( \{0, 0.44, 0.89\} \).