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    So the question is, "solve for the following in the interval shown:
    Sin 2X = Sin X (0<X<360) (< = less than or equal)

    Now I was able to bring this down to SinX = 1/2 , leaving me with 60 degrees and 300 degrees for values of X, but the mark scheme is showing two more values?? Can someone explain this please?
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    (Original post by ludd-sama)
    So the question is, "solve for the following in the interval shown:
    Sin 2X = Sin X (0<X<360) (< = less than or equal)

    Now I was able to bring this down to SinX = 1/2 , leaving me with 60 degrees and 300 degrees for values of X, but the mark scheme is showing two more values?? Can someone explain this please?
    Sub X = 0

    i.e. where 2X = X

    :cute:
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    You mean sinx=sin(2x) thus sinx(2cosx-1)=0 thus sinx=0 or cosx=1/2 (not sin as you wrote)
    the other two values are from sinx=0
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    Sin2x = SinX
    2SinXCosX - Sinx = 0
    Sinx(2CosX-1) = 0
    Sinx=0 CosX= 1/2

    I am pretty sure you are able to do it like that..
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    Ha you beat me to it...
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    (Original post by Vesniep)
    You mean sinx=sin(2x) thus sinx(2cosx-1)=0 thus sinx=0 or cosx=1/2 (not sin as you wrote)
    the other two values are from sinx=0
    yh I meant cos X sorry, but yh thanks I see exactly where I went wrong now, it didn't occur to me that I could make the equation equal 0.
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    (Original post by Indigo.Brownhall)
    Sin2x = SinX
    2SinXCosX - Sinx = 0
    Sinx(2CosX-1) = 0
    Sinx=0 CosX= 1/2

    I am pretty sure you are able to do it like that..
    yep, really helpful, thanks a lot
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    I actually have one more question if anyone would be willing to help.

    "3cosX - sin(X/2) - 1 = 0" (0<X<720)

    I was able to factorise it down to (9cosX - 1)(2cosX-1) leaving me with The values of X being: 60, 83.6, 276.4, 300, 420, 660, 636.4 and 443.6.
    The issue is, the mark scheme only contained 4 of these (60, 300, 443.6, 636.4). Is there a reason why only half of the answers I got were accepted, I don't think there was a fault in my working out?
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    (Original post by ludd-sama)
    I actually have one more question if anyone would be willing to help.

    "3cosX - sin(X/2) - 1 = 0" (0<X<720)

    I was able to factorise it down to (9cosX - 1)(2cosX-1) leaving me with The values of X being: 60, 83.6, 276.4, 300, 420, 660, 636.4 and 443.6.
    The issue is, the mark scheme only contained 4 of these (60, 300, 443.6, 636.4). Is there a reason why only half of the answers I got were accepted, I don't think there was a fault in my working out?
    You can check each of your answers by putting them into the equation and testing which work.
    E.g. 83.6 doesn't work.
    If you show your working someone can help yo find your error


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    (Original post by gdunne42)
    You can check each of your answers by putting them into the equation and testing which work.
    E.g. 83.6 doesn't work.
    If you show your working someone can help yo find your error


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    83.6 doesn't work (bear in mind this is rounded to 3dp)? when I put in the arccos of 1/9 I get 83.6. as for the working..
    I made the equation equal sin(X/2) and plugged in the relevant half angle formula for sin(X/2), then squared both sides, leaving me with a quadratic which I factorised to (9cosX -1)(2cosX -1) which I'm sure is right. obviously this left me with cosX=1/9 and cosX=1/2, and then I just used my calculator and CAST method to get out all the values for X (which all checked out on the calculator)
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    (Original post by ludd-sama)
    83.6 doesn't work (bear in mind this is rounded to 3dp)? when I put in the arccos of 1/9 I get 83.6. as for the working..
    I made the equation equal sin(X/2) and plugged in the relevant half angle formula for sin(X/2), then squared both sides, leaving me with a quadratic which I factorised to (9cosX -1)(2cosX -1) which I'm sure is right. obviously this left me with cosX=1/9 and cosX=1/2, and then I just used my calculator and CAST method to get out all the values for X (which all checked out on the calculator)
    3cos(83.6)-sin(83.6/2)-1 does not equal anything close to 0

    Squaring both sides of an equation can easily introduce solutions that are not valid for the original equation.

    Consider
    Cos(X)=0.5 (2 solutions 0-360)
    Cos^2(X)=0.25
    So cos(X)=root(0.25) = plus or minus 0.5 (4 solutions 0-360)

    Using the double angle formula to rewrite 3cos(X) in terms of sin(X/2) is an alternative approach

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    (Original post by Indigo.Brownhall)
    Ha you beat me to it...
    unlucky brownhole
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    I think you're dividing by sinx or dividing by trig terms and therefore losing answers. You always want to factorise and not just divide. What I mean is for example if you have
     \cos x \sin x +\frac{1}{2} \cos x = 0 .
    You should factor out the cosx and not just divide by cosx as you will lie answers.
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    (Original post by Alexion)
    Sub X = 0

    i.e. where 2X = X

    :cute:
    sin isn't injective. That's not the only solution.
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    (Original post by morgan8002)
    sin isn't injective. That's not the only solution.
    I know it's not the only solution - OP was just missing the zero/2pi solution from his answer.
 
 
 
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