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    Hi, I need help with this question if you can please

    Find the values of x in (0,360 degrees) for which 7cosx+sinx=5 giving each answer to 1 decimal place.

    I don't understand the (0,360 degrees) bit of this question
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    (Original post by TheQueen1986)
    Hi, I need help with this question if you can please

    Find the values of x in (0,360 degrees) for which 7cosx+sinx=5 giving each answer to 1 decimal place.

    I don't understand the (0,360 degrees) bit of this question
    There are infinitely many x that solve your equation. They are asking only for those solutions that are in the range

    0 < x < 360 degrees.
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    Oh I see! Thanks
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    Damn, I need help once again on this one.

    I did as follows and I don't end up with the right answers. I know this because when I substitute the answers back into the original equation, they don't equal to 5.

    7cosx+sinx=5
    49cos(squared)x+sin(squared)x=25
    49(1-sin(squared)x)+sin(squared)x=25
    49-49sin(squared)x+sin(squared)x=25
    -48sin(squared)x=-24
    sin(squared)x=0.5
    sinx=+ or - square root of 0.5
    x=45 degrees, 315 degrees.
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    [QUOTE=TheQueen1986]Damn, I need help once again on this one.

    I did as follows and I don't end up with the right answers. I know this because when I substitute the answers back into the original equation, they don't equal to 5.

    7cosx+sinx=5
    49cos(squared)x+sin(squared)x=25
    QUOTE]
    Your error occurs in the second line.
    if you square both sides you will have a 14cosxsinx term.
    compare 7cosx+sinx to Rcos(x-A) (*)
    so 7cosx+sinx=RcosxcosA+RsinxsinA
    which leads to
    RcosA=7 (1)
    Rsin A=1 (2)

    dividing (2) by (1) gives sinA/cosA=1/7 so Tan A=1/7
    squaring (1) and (2) and adding gives
    R^2(cos^2A+sin^2A)=49+1
    so R^2=50
    then to finish off use these values for R and A in (*)
    to get
    Rcos(x-A)=5
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    oh yeah, I forgot, this is like the common type they ask in exams isn't it? Thanks for your help!
 
 
 
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