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Math Help - Solve equation

  1. #1
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    Solve equation

    Solve the equation over the intervals [0,2pi)

    3cosx + 3 = 2sin²x

    2sin²x - 3cosx - 3 = 0 Put it all on one side

    1-2cos²x - 3cosx - 3 = 0 Used the pythagorean identity that Sin²x + Cos²x = 1

    -2cos²x - 3cosx - 2 = 0 Combined the integers

    From here i was confused because it doesn't factor into two binomials due to -cos²x being negative. So.. i made -cos²x positive using the even/odd identity. I am not sure if that is right.

    (2cosx + 1)(cosx - 2) = 0 Used even/odd identity

    So cosx = -1/2, 2

    2pi/3, 4pi/3 My answer, didn't know what to do with the 2

    Is this correct?
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  2. #2
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    Quote Originally Posted by brandonb122 View Post
    Solve the equation over the intervals [0,2pi)

    3cosx + 3 = 2sin²x

    2sin²x - 3cosx - 3 = 0 Put it all on one side

    2-2cos²x - 3cosx - 3 = 0 Used the pythagorean identity that Sin²x + Cos²x = 1

    -2cos²x - 3cosx - 2 = 0 Combined the integers

    From here i was confused because it doesn't factor into two binomials due to -cos²x being negative. So.. i made -cos²x positive using the even/odd identity. I am not sure if that is right.

    (2cosx + 1)(cosx - 2) = 0 Used even/odd identity

    So cosx = -1/2, 2

    2pi/3, 4pi/3 My answer, didn't know what to do with the 2

    Is this correct?
    See my note in blue. You need to expand the 2 across the 1 as well as cos^2(x)

    -2\cos^2(x)-3\cos(x) -1 = 0 \implies 2\cos^2(x) + 3\cos(x)+1 = 0


    edit: b^2-4ac = 1 so this factors
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