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    Can someone come lease tell me where I have gone wrong here because this is obviously not going to work out.
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    (Original post by Hannahkisley)
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    Can someone come lease tell me where I have gone wrong here because this is obviously not going to work out.
    you can replace 1 - cos2x with sin2x

    then you have sin2x = 2sinx

    so sin2x - 2sinx = 0

    now you should factorise the LHS
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    (Original post by Hannahkisley)
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    Can someone come lease tell me where I have gone wrong here because this is obviously not going to work out.
    If the original question was

    \sin^2 x = 2\sin x

    then you should factorise instead of dividing by \sin x. Dividing an equation by a trig function will often mean you lose solutions so it should be avoided.

    For future posts please try to make your images clearer and post them the correct way up.
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    sin^2(x) is not 2sin(x). Rearrange the identity cos^2(x) + sin^2(x) = 1.
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    (Original post by Edgemaster)
    sin^2(x) is not 2sin(x). Rearrange the identity cos^2(x) + sin^2(x) = 1.
    That’s what I did and I got sin^2 x = 1-cos^2 x so I replaced 1-cos^2 x in the initial equation with sin^2 x to get sin^2 x = sin x
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