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    Solve the equation..

    cos x - (rt3)sin x = rt2

    between 0 and 360.
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    (Original post by jonnymcc2003)
    Solve the equation..

    cos x - (rt3)sin x = rt2

    between 0 and 360.


    Whats rt3 and rt2?
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    (Original post by BlueAngel)
    Whats rt3 and rt2?
    square root
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    (Original post by BlueAngel)
    Whats rt3 and rt2?
    Square root of 2 etc
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    (Original post by jonnymcc2003)
    Solve the equation..

    cos x - (rt3)sin x = rt2

    between 0 and 360.

    Thats a hard one that.
    rt3/2 is Sin 60 or Cos 30

    Oh I dont know.
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    I get there is no solution...so I think I might have done it wrong lol

    but what I basically did is put the sin on the other side, then square both sides, then make cos^2x = 1-sin^2x so now you've only got sin^2x's sinx's and units, so you've got a quadratic, then make a substituion of u=sin x and solve
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    i dunno if there are more solutions, but you get the idea
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    Same method as katie heskins

    CosX - √3SinX = √2
    CosX = √2 + √3Sinx
    Cos²X = 2 + 2√6Sinx + Sin²X
    (1 - Sin²X) = 2 + 2√6Sinx + Sin²X
    2Sin²X + 2√6Sinx + 1 = 0

    Then solve using the quadratic formula.

    I got sinX = -0.225 or nothing as the other answer gives an answer smaller than -1.

    So X = 347 or 193.0

    don't know if it's right though.
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    hmm

    i used rcos(theta + alpha)

    r=root(1+3)=2

    alpha = tan^-1(-root3/1)

    so you get

    2cos(theta-60)=root(2)

    etc
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    For a P2 student I daresay Kitzens method is the one that the exam board would have in mind.
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    Just to add to Kikzens solution. There of course is a second solution. Remember that cosx = cos(-x), so
    cos(x+60) = 1/sqrt(2)
    x+60 = 45, -45

    Gives, x = 255 and 345.
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    Also, Katie and Mysticisms method works, but only finds one solution. Because you have squared, you get another solution which will not work.
 
 
 
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