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Sin3x - sinx = 0 (for x greater than 0 but less than 360)

  1. Feb 2, 2008 #1
    Hi folks, can anybody help me. I would like to solve the following equation: ""sin3x - sinx = 0 (for x greater than 0 but less than 360) and come up with angles for x or 2x. We are doing Trigonometrical identities which i have only just come across. Got as far as "sin(2x + x) - sinx = 0" as we think we need to get a 2x or x and sin(2x + x) is also an identity that can be changed, but not sure if this is the right direction.

    Many thanks folks
     
  2. jcsd
  3. Feb 2, 2008 #2
    Can you write the x (in the second term) in terms of 2x and x like you did for 3x?
     
  4. Feb 2, 2008 #3
    I believe so as this gives "sin(a + b)" (or in my case sin(2x + x)) and this is equal (or can then be substituted with the identity) "sinacosb + cosasinb", BUT i really don't know if i'm heading the correct way as i've only just this week come across trigonometrical identities so it could be totally different. I know the last equation i did gave me a double angle (2x) which gave 2 angles within 360 degrees. Any help you can give would be most appreciated.
     
  5. Feb 2, 2008 #4
    I meant this: sin(2x+x) - sin(2x-x) = 0.
     
  6. Feb 2, 2008 #5
    Well if you want to try it this way let a=2x and b=x, then use that same formula again on the terms that have 2x's in them. You only want sines from then on....
    But it's a lot of algebra
     
  7. Feb 2, 2008 #6
    I don't think so as sin(2x+x) could be a backwards step from sin3x but with the "- sinx" there is only one x there so not sure where the 2x and x would come from.
     
  8. Feb 2, 2008 #7

    VietDao29

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    Homework Helper

    There are actually 3 ways to solve the above problem.

    1. The first way, the most straightforward, and require the most calculation is to trace 3x down to x, by using Triple-Angle Formulae: sin(3x) = 3 sin(x) - 4sin3(x) (You can arrive to this formula by using the Sum-Angle Identity twice). So, your equation becomes a cubic equation in sin(x), which is pretty easy to solve. :)

    2. The second way, the easiest way, is to use the properties of sin function:

    [tex]\sin \alpha = \sin \beta[/tex]

    [tex]\Leftrightarrow \left[ \begin{array}{lcr} \alpha & = & \beta + k 360 ^ o \\ \alpha & = & 180 ^ o - \beta + k' 360 ^ o \end{array} \right.[/tex], k, and k' are both integers.

    One can isolate sin(x) to the other side of the equation: sin(3x) = sin(x), and use method mentioned above. Then choose, k, and k' wisely so that your solution is on the interval [0, 360]

    3. The 3rd way, the final one, is to use the Sum-To-Product Identities:
    (You can arrive to this Itentity by using neutrino's hint)
    [tex]\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]

    Hopefully, you can go from here. :)

    You can pick up one of the 3 ways mentioned above, or try all 3, and compare the result. :)
     
    Last edited: Feb 2, 2008
  9. Feb 2, 2008 #8
    I'm not sure if thats at all the correct way but a friend of mine started by using sin(a + b) as this was an identity that could be replaced. These trig identities are all new to me and not the best at algebra either
     
  10. Feb 2, 2008 #9
    Thanks for your help. Its all very confusing as i'm quite new to this level of maths. I will have a look through.
     
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