## Algebra 2 (1st Edition)

$t=0,12,24$ for low tide and $t=6, 18$ for high tide
From the previous part (a), we have $d= -6.5 \cos (\dfrac{\pi}{6})t+10$ Here, Low Tide $d=3.5$ $3.5= -6.5 \cos (\dfrac{\pi}{6})t+10 \implies \cos (\dfrac{\pi}{6})t=1$ or, $2 \pi n=\dfrac{\pi}{6}t \implies 12n=t$ This gives: $t=0,12,24$ Here, High Tide $d=16.5$ $16.5= -6.5 \cos (\dfrac{\pi}{6})t+10 \implies \cos (\dfrac{\pi}{6})t=-1$ or, $\pi+2 \pi n=\dfrac{\pi}{6}t \implies 6+12n=t$ This gives: $t=6, 18$