only a rough sketch of a possible solution (I haven't much time yet )
1. The function w(t) = 1.1 - sin((pi/6)*t) doesn't have any zeros.
2. 1. April 1990 correspond to w(23) = 8/5 (1.6*10^6 gallons)
3. After that date you deal with a new function:
a(t) = (1.1 - 0.25) - sin((pi/6)*t)
4. This function has a lot of zeros. Calculate the zero directly after t = 23
5. Good luck!