Math 12

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trig question
2sin^2x+sinx-6=0
how to solve for x when the
restriction is x is greater than or equal to zero but less than 2pi

  • Math 12 -

    this is a quadratic, it might be easier to see if you let
    sinx = t, then we have

    2t^2 + t - 6 = 0 which factors to
    (2t - 3)(t + 2) = 0
    t = 3/2 or t = -2

    so sinx = 3/2 or sinx = -2

    both cases are not possible since sinx must be between -1 and +1

    so your equation has no solution.

  • Math 12 -

    let y = sin x
    the equation then becomes:
    2y^2 + y - 6 = 0
    then factor:
    (2y - 3)(y + 2) = 0
    y = 3/2 and y = -2
    therefore:
    sin x = 3/2, x = 2*pi*n + arcsin 3/2 ; and
    sin x = -2 , x = 2*pi*n + arcsin (-2) , where n = integer
    *but note that arcsin 3/2 and arcsin (-2) are undefined.

    so there,, :)

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