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Solving A Trignometric Equation

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find all solutions to the equation in the interval [0,2[tex]\pi[/tex]] algebraically.

    cos2x(2cos+1)=0

    2. Relevant equations

    NA

    3. The attempt at a solution

    This is what I've done, but I don't think it's right. I set cos2x=0 and 2cos+1=0. Since cos2x is a multi-angle, I let t=2x and rewrote it as cost=0 That means t equals [tex]\pi[/tex]\2+n[tex]\pi[/tex].

    That also means 2x=[tex]\pi[/tex]\2+n[tex]\pi[/tex]. After dividing out the 2, I get x=[tex]\pi[/tex]\4+n[tex]\pi[/tex]\2.

    I get the answers [tex]\pi[/tex]\4, 3[tex]\pi[/tex]\4, 5[tex]\pi[/tex]\4, and 7[tex]\pi[/tex]\4. This doesn't seem right to me. Shouldn't I only get two answers.

    The other one seems easier. I got cosx by itself and it is cosx=-1\2. That means x is equal to 2[tex]\pi[/tex]\3 and 4[tex]\pi[/tex]\3.

    Am I doing this correctly?

    Thanks!
     
  2. jcsd
  3. Oct 9, 2009 #2
    The pi symbols aren't supposed to be raised up like that. It's just pi; sorry about that.
     
  4. Oct 9, 2009 #3
    cosx=0 has two solutions, but you have four solutions to cos2x=0. That 2 multiplied with the x in cos2x makes its period only half as long as that of cosx, going from a period of 2[itex]\pi[/itex] to [itex]\pi[/itex]
    cosx=0 has just two solutions in [0, [itex]\pi[/itex]], so cos2x=0 should have twice as many solutions, because it now goes through two periods in [0, [itex]\pi[/itex]].

    Hopefully this doesn't make it more complicated for you. You can also look at the graph of cos2x to see that there are four solutions in [0, [itex]\pi[/itex]].


    Instead of the [ tex ] tags, you can use [ itex ] where the LaTeX fits better within other text like you have. Also, your division lines should be going the other way. For example, [itex]\pi[/itex]/4
     
  5. Oct 9, 2009 #4
    Thank you for you reply! I plugged in the equations separately, and then as the original equation. It really helped visualize what was going on and I could double check using the trace function. I did not think to try this before. It all makes sense now!

    I'll try to clean up my tags and symbols; thanks for the tips.
     
  6. Oct 10, 2009 #5

    HallsofIvy

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    ?? What does "2cos+1" mean? Is that supposed to be "2cos(x)+ 1"?
    If the problem is to solve cos(2x)(2cos(x)+ 1)= 0 then, yes, either cos(2x)= 0 or 2cos(x)+ 1= 0. cosine is 0, in the interval [itex][0, 2\pi][/itex] only for 0 and [itex]2\pi[/itex] so you have 2x= 0 and [itex]2x= 2\pi[/itex] as roots.

    2cos(x)+ 1= 0 gives cos(x)= -1/2. Think about dividing an equilateral triangle in half.
     
    Last edited: Oct 11, 2009
  7. Oct 10, 2009 #6
    It's supposed to be 2cos(x)+1. I'm sorry, I mistyped it. I calculated the problem correctly and Bohrok helped me understand why x has twice as many solutions for cos(2x)=0.
     
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