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Simple trigonometry problem

  1. Feb 1, 2010 #1
    Solve [tex]3\cos\frac{3x}{2}=\cos\frac{x}{2} [/tex] I am interested to know how to go about solving such an equation where the coefficient of cos(x) is not equal to 1. I know how to solve [tex] \cos\theta = \cos\alpha, \mbox{ it is just} \theta = 2n\pi +/-\alpha [/tex] I teach various maths subject to myself, then I realised I didn't know how to solve the above equation. Thanks.
     
  2. jcsd
  3. Feb 1, 2010 #2
    Well, [tex] \cos \frac{3x}{2} = \cos (x + \frac{x}{2}) [/tex]

    Therefore, you get [tex] x = (2n + 1)\pi \ for \ n = ...-3, -2, -1,0,1,2,3,... [/tex] as a freebie.
     
  4. Feb 1, 2010 #3
    Surely the non-trivial solutions are more interesting.

    You can always transform multiple angles to powers of trigonometric functions (and vice versa).
    For example with de Moivre
    [tex]\cos ny=\Re\left(\cos y+i\sin y\right)^n[/tex]
    [tex]\cos 3y=\cos^3y-3\cos y\sin^2y=\cos^3y-3\cosy(1-\cos^2y)=\cos y(4\cos^2y-3)=\cos y(2\cos 2y-1)[/tex]

    Since here on angle is three times the other, you can use y=x/2 and get
    [tex]3\cos 3y=\cos y[/tex]
    [tex]3\cos y(2\cos(2y)-1)=\cos y[/tex]
    and therefore either
    [tex]x=\pm\frac{\pi}{4}+k\pi[/tex]
    or
    [tex]x=\pm\frac14\arccos\frac23+k\frac{\pi}{2}[/tex]
     
  5. Feb 3, 2010 #4
    Thanks for the time you spent on working that out. I didn't know of De Moivre's. I worked it out afterwards and get [tex] 2y=\pm \arccos(\frac{2}{3})+ 2n\pi[/tex]. And y=x/2, I couldn't [tex] x=\pm\frac{\pi}{4}+k\frac{\pi}{2}[/tex], which is what you got. According to my graphing calculator the graph crosses the x-axis at [tex]\arccos(\frac{2}{3})[/tex] which looks close to pi/4 but is not. It crosses the x-axis again close to pi.
     
  6. Feb 3, 2010 #5
    Not sure what you mean.

    Both 1/4arccos 2/3 and pi/4 are solutions.
     
  7. Feb 3, 2010 #6
    I get a principle angle of arccos(2/3) = 2y and -arccos(2/3) is another one , right. Maybe I have already gone wrong. If I saw the steps you used I could follow you. [tex]
    3\cosy(2\cos(2y)-1)=cosy, \mbox{therefore}\\
    3(2\cos(2y)-1)=1, \\
    2\cos(2y)-1=\frac{1}{3}, \\
    2\cos(2y)=4/3, \\
    \cos(2y)=\frac{2}{3} [/tex]
     
  8. Feb 3, 2010 #7
    The equation is
    [tex]3\cos y(2\cos(2y)-1)=\cos y[/tex]
    Now either cos(y) is zero (which gives you pi/4 for x) or you can divide by cos(y) and continue to get x=arccos(2/3)/4
     
  9. Feb 3, 2010 #8
  10. Feb 4, 2010 #9
    If y = x/2 then x = 2y, so if the solutions for y are

    [tex] y = \frac {\pi} {2} + k \pi[/tex]

    or

    [tex] y = \pm \frac {1} {2} \arccos \frac {2}{3} + k \pi [/tex]

    then the solutions for x will be

    [tex] x = (2k + 1) \pi[/tex]

    or

    [tex] x = \pm\arccos \frac {2}{3} + 2 k \pi [/tex]
     
    Last edited: Feb 4, 2010
  11. Feb 6, 2010 #10
    Yes, that is what I finally finished up with, willem2.Thanks willem2.
     
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