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anonymous

  • 5 years ago

dy/dx = -2sinx + 2cos2x How do I find its critical points?

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  1. myininaya
    • 5 years ago
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    set =0

  2. anonymous
    • 5 years ago
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    Yes I know this step, its just that I get a cos on one side and a sin on the other side. I don't know how to switch them to the same trig function to solve for x

  3. anonymous
    • 5 years ago
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    double angle cos(2x) = 1 - 2sin(x)^2

  4. anonymous
    • 5 years ago
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    ok so the I did the double angle and that gives me: 2sinx = 2(cos^2x - sin^2x) How did you get 1 -2sin(x)^2?

  5. anonymous
    • 5 years ago
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    cos(2x) = 1 - 2sin(x)^2 it's a rule you should have 2sinx = 2 - 2sin^2x

  6. anonymous
    • 5 years ago
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    it's called the double angle formula

  7. anonymous
    • 5 years ago
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    Funny, my book says: Cos2A = cos^2A - sin^2A

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