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Math Help - two tricky trig problems

  1. #1
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    two tricky trig problems

    Show All Work Please

    1) solve: cos2x-cos6x=0

    and

    2) The number of hours daylight in Detroit on day 't' ( with t = 0 on January 1st) of a non-leap year is given by:
    d(t)=3sin[(2π/365)(t-80)]+12

    a) What days have exactly 11 hrs. of daylight
    b)What day has a maximum amount of daylight

    if you thought those were easy try

    Solve: X (to the 3/8) - (square root of negative one)(square root of 3)=3
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  2. #2
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    Quote Originally Posted by rutheada000 View Post
    Show All Work Please

    1) solve: cos2x-cos6x=0

    [snip]
    => cos(2x) = cos(6x).

    Note that the general solution to cos(A) = cos(B) is:

    A = B + 2 n pi and A = -B + 2 n pi

    where n is an integer .....
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