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    How do I solve: 3cosx+4sinx+3=0 Where x is between 0 & 360

    I'm not sure how to begin !

    Thanks,
    Ross
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    try expressing it in the form of Rcos(x+/-a) or Rsin(x+/-a)
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    I think iteration is the only way to solve this kind of question. I'm in AS so I don't know that how iteration is applied, but you may search on google for it.
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    ive just finished this chapter in C2 last week and i think it wud be helpful if u used the formula: sinx/cosx = tanx

    so it would be 4/3tanx + 3=0
    then make tanx the subject and find tan-1 in ur calculator and then u can use the quadrant rule or the graph to help u find the angles
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    (Original post by cj_134)
    try expressing it in the form of Rcos(x+/-a) or Rsin(x+/-a)
    Zishi gave you the correct answer for people with C3 knowledge.

    Another approach for those with a little more knowledge is to use a half tangent substitution.
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    (Original post by Rossblunsden)
    How do I solve: 3cosx+4sinx+3=0 Where x is between 0 & 360

    I'm not sure how to begin !

    Thanks,
    Ross
    THere is more way to solve this equation.
    For example you can solve it graphically.

    It seems most simple for me to use trigonometric identity of
    sin^2x+cos^2x=1
    and from this change sinx with
    sinx=\sqrt{1-cos^2x}
    And arrange the equation
    4\sqrt{1-cos^2x}=3(1-cosx)
    Take square both side and solve for cosx


    Another method is the use of another tigonometric identity.
    That is the addition rule
    sin(A+x)=sinAcosx+cosAsinx
    We don't know sinA and cosA but we know from the equation that
    the ratio of them is 3/4 that is
    sinA=3t
    cosA=4t
    Taking square both equation and adding them
    sin^2A+cos^2A=25t^2
    1=25t^2
    t=\frac{1}{5}
    So sinA=3/5 cosA=4/5 and tanA=3/4 -->A
    Divide the original equation by 5 and substitute
    \frac{3}{5}cosx+\frac{4}{5}sinx+  \frac{3}{5}=0
    sinAcosx+cosAsinx=-\frac{3}{5}
    sin(A+x)=-\frac{3}{5}


    THe third method maybe that arranging the equation you use the identity of
    sin^2\frac{x}{2}=\frac{1-cosx}{2} and
    sinx=2sin\frac{x}{2}cos\frac{x}{  2}
    THe equation
    4sinx=3(1-cosx)
    8sin\frac{x}{2}cos\frac{x}{2}=6s  in^2\frac{x}{2}
    Arranging to 0 and taking out sinx/2 as factor it gives the solutions


    The 4th method would be writing up sine and cosine with t=tanx/2
    t=tan\frac{x}{2}
    sin^2\frac{x}{2}+cos^2\frac{x}{2  }=1
    dividing by cosine squared
    tan^2\frac{x}{2}+1=\frac{1}{cos^  2\frac{x}{2}}
    \frac{1}{1+t^2}=cos^2\frac{x}{2}
    sin^2\frac{x}{2}=1-cos^2\frac{x}{2}=\frac{t^2}{1+t^  2}
    So
    cosx=cos^2\frac{x}{2}-sin^2\frac{x}{2}=\frac{1-t^2}{1+t^2}
    sin^2x=4sin^2\frac{x}{2}\cdot cos^2\frac{x}{2}=\frac{4t^2}{(1+  t^2)^2}
    that is
    sinx=\frac{2t}{1+t^2}
    Substituting them to the original equation we get a quadratic for t.


    The fifth method maybe if use the identity of
    \pm \sqrt{1+tan^2x}=\frac{1}{cosx}
    Arranging the equation and dividing by cosx (cosx=0 would be examined as solution)
    3cosx+4sinx=-3
    3+4tanx=\frac{-3}{cosx}=-3\sqrt{1+tan^2x}
    So we've got a quadratic equation for tanx


    Another method would be using complex valued, so
    exponential functions.
    cosx=cosh(ix)=\frac{1}{2}(e^{ix}  +e^{-ix})
    sinx=\frac{1}{i}sinh(ix)=\frac{1  }{2i}(e^{ix}-e^{-ix})
    This give quadratic for e^(ix).
 
 
 
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