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Math Help - [SOLVED] Sin, cos! Urgent!

  1. #1
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    [SOLVED] Sin, cos! Urgent!

    Barron Page 77

    Solve for 2sinx+ cos2x=2sin^2x-1 for 0< and equals x < and equals 2 \pi?

    Thanks in advance! Urgent!!
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  2. #2
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    Hint: \cos 2x = 1 - 2\sin^2 x

    Arrange your equation and you will have it in the form of: ay^2 + by + c = 0

    Then you can use the quadratic formula.
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  3. #3
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    Ok thanks.

    I got 4sin^2x-sinx+2=0
    do i just solve x? I don't have to consider sin?
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    You solve the quadratic taking sin(x) as the variable and then you find x with
    x = arcsin(sin(x)) there can be 2 solutions.
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    How do you when to use arcsin or when to use sin?
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    Senior Member vincisonfire's Avatar
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    Suppose
    a\cdot sin(x)^2 + b \cdot sin(x) + c = 0
     sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
     x = arcsin(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a})
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