Trigonometry..

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During high tide the water depth in a harbour is 22 m and during low tide it is 10m(Assume a 12h cycle).
Calculate the times at which the water level is at 18m during the first 24 hours.

My solution:
first I found the cos equation: H(t)=-6cos(π/6t)+16
then..
Let π/6t=Θ
18=-6cosΘ+16
18-16=-6cosΘ
Θ=1.230959417


Then I don't know what's next....

  • Trigonometry.. -

    H(t)=-6\cos(\frac{\pi}{6} t)+16

  • Trigonometry.. -

    cosΘ < 0 in QII and QIII
    arccos(-1/3) = 1.91 = pi-1.23
    or pi+1.23 = 4.37

    So, t = (6/pi) * (1.91 or 4.37)
    t = 3.64 or 8.35 hours

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