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I need help with a question on FP2 complex numbers :( Watch

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    I'm just trying to do part 2 of a question on complex numbers. The first part asked to show that sin^5 x = \frac{1}{16}(sin^5 x -5sin^3 x + 10sinx). I did that without a problem, but now it says "hence solve, in the interval 0 <= x <= 2pi, sin5x - 5sin3x + 6sinx = 0. My problem is that I don't even remember doing this type of questions, I did things like z^4 + 16i = 0, but I just don't know how to do this one. Can anyone start me off?
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    Assuming that the first part was
    \displaystyle \sin^5 x=\frac{1}{16}\left (\sin 5x-5\sin 3x+10\sin x\right )
    The question is
    \displaystyle \sin 5x-5\sin 3x+6\sin x=(\sin 5x-5\sin 3x+10\sin x)-4\sin x=
    \displaystyle =16\sin^5 x - 4\sin x=0
    with factorization you will get the solutions.
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    (Original post by ztibor)
    Assuming that the first part was
    \displaystyle \sin^5 x=\frac{1}{16}\left (\sin 5x-5\sin 3x+10\sin x\right )
    The question is
    \displaystyle \sin 5x-5\sin 3x+6\sin x=(\sin 5x-5\sin 3x+10\sin x)-4\sin x=
    \displaystyle =16\sin^5 x - 4\sin x=0
    with factorization you will get the solutions.
    Thanks a lot man, it really helps. I did:  16sin^5 x - 4sinx = sinx (4sin^2 +2)(4sin^2 x -2). I already know that x = 0, Pi/4, Pi, 5Pi/4 and 2Pi, but what about sinx =\sqrt\frac{-1}{2}? I'm not sure how to approach this part, it looks like sinx = \frac{i}{2}, how can i solve it?
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    (Original post by Draggy)
    Thanks a lot man, it really helps. I did:  16sin^5 x - 4sinx = sinx (4sin^2 +2)(4sin^2 x -2). I already know that x = 0, Pi/4, Pi, 5Pi/4 and 2Pi, but what about sinx =\sqrt\frac{-1}{2}? I'm not sure how to approach this part, it looks like sinx = \frac{i}{2}, how can i solve it?
    Since you are told that x is in the range 0 to 2pi, x must be a real angle, so that particular factor has no zeros!
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    (Original post by Draggy)
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    sinx = \frac{i}{2} has no real solutions.


    Also, you are missing some solutions to the factor 4sin^2 x -2. There are 4 of them in the given interval.
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    (Original post by davros)
    Since you are told that x is in the range 0 to 2pi, x must be a real angle, so that particular factor has no zeros!
    Right, so are these five angles the only ones to be found?
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    (Original post by Draggy)
    Right, so are these five angles the only ones to be found?
    See my previous post - there are 7 in all.
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    (Original post by ghostwalker)
    See my previous post - there are 7 in all.
    Ok, from sinx = 0 we get 0, Pi and 2 Pi. I think I did the other one wrong, sinx = \frac{1}{\sqrt2} gives Pi/4 and 3Pi/4. I'm looking at the CAST diagram and can't find the other two.
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    (Original post by Draggy)
    Ok, from sinx = 0 we get 0, Pi and 2 Pi. I think I did the other one wrong, sinx = \frac{1}{\sqrt2} gives Pi/4 and 3Pi/4. I'm looking at the CAST diagram and can't find the other two.
    sinx = \pm\frac{1}{\sqrt2}
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    (Original post by ghostwalker)
    sinx = \pm\frac{1}{\sqrt2}
    Ah, completely forgot about that... :rolleyes: The other two angles are 5Pi/4 and 7Pi/4. Thanks a lot for help guys, I feel like I understand all of this now.
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    (Original post by Draggy)
    ...
    FYI.
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