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Math Help - Solving trig

  1. #1
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    Solving trig

    hi. ive been given this equation
    12cosx -4 sinx=7 solve between 0<x<360 degrees.
    ive already done the harmonic form for f(x)= 12cosx -4 sinx in the form
    = Rcos(x + a) and got square root 160cos(x + 18.43)
    im just not sure how to solve it
    thanks.
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  2. #2
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    Try t method substitution

    sin@ = 2t/1 + t^2
    cos@ = 1-t^2/1+t^2

    12cosx -4 sinx=7 solve between 0<x<360 degrees.

    12(1-t^2/1+t^2) - 4(2t/1+t^2) = 7
    (12 - 12t^2/1+t^2)- (8t/1 + t^2) = 7
    12- 12t^2 - 8t = 7 + 7t^2
    12 - 19t^2 - 8t - 7 = 0
    5 - 19t^2 - 8t =0
    -5 + 19t^2 + 8t = 0

    -8 +- sqrt 64 - 4x19 x -5
    -------------------------
    38

    t = 0.344 or -0.765

    tan@/2 = 0.344
    @/2 = 18 degrees, 59 minutes
    @ = 37 degrees, 58 minutes

    or 180 + 37.97 = 217degrees, 58 minutes

    tan@/2 = 0.765
    @/2 37.42
    but thingo is negative - second and 4th quadrant
    therefore, @/2 = 142.58 or 322.58

    @ = 285.16 or 645.16
    @ = 285.16 (within the domain)

    Rechecking this... It's not that accurate. Sorry, I'm new to trigonometry.
    ----------------------------------------------

    Using the auxiliary angle method:

    12cosx -4 sinx=7 solve between 0<x<360 degrees.

    acos@ - bsin@ = Acos(@+a)

    tana = b/a
    tana = 4/12
    a = 18 degress, 26 minutes

    A = sqrt a^2 + b^2
    = sqrt 144 + 16
    = sqrt 160

    sqrt 160 cos (@ + 18degrees,26 minutes) = 7
    cos (@ + 18 degrees, 26 minutes) = 7/sqrt 160
    cos 56 degrees, 24 minutes OR 303 degrees, 36 minutes

    @ + 18 degrees, 26 mins = 56 deg, 24 mins
    @ = 37 degrees, 58 mins

    @ + 18 degrees, 26 mins = 303 degrees, 36 mins
    @ = 285 degrees, 10 mins

    Don't know if its right, but please don't assume it is!
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  3. #3
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    cheers mate , thats brilliant
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