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Solving Trig Identity?

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Need some help finding all solutions for x....

    csc^2((x)/(2)) = 2secx




    3. The attempt at a solution
    Not sure what kind of approach to take but:

    1/ sin^2(x/2) = 2/ cos x

    From here Not sure what to do i tried cross multiplying and got cos x = 2sin^2(x/2) but got no idea from here.... please help
     
  2. jcsd
  3. Dec 8, 2012 #2
    Does anyone have an explanation for this solution?? someone just sent it:

    csc^2(x/2)=2secx
    2/(1-cosx)=2secx
    2=2secx(1-cosx)
    1=secx-1
    2=secx
    π/3+2πk=x
    5π/3+2πk=x

    However i do think hes missing a squared? Can anyone explain this formula?
     
  4. Dec 8, 2012 #3

    rock.freak667

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    Is your problem

    cosec2(x/2) = secx

    or

    ½cosec2(x)=secx


    If it is the first one, then your double angle formulas for cos2x will help you out here greatly.
     
  5. Dec 9, 2012 #4

    Mark44

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    The line above is incorrect on the left side. It should be
    2/(1 - cos2(x/2)) = 2sec(x)
    To solve this equation, convert everything into terms involving cosine. You need the double angle formula to convert cos(x) into cos(2 * x/2).

    BTW, your title is misleading. You do not "solve" an identity - you prove that an equation is identically true (true for all or most values of the variable). What this seems to be is an equation to solve for specific values of x.
     
  6. Dec 9, 2012 #5

    SammyS

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    I would be inclined to go from where you are here
    cos(x) = 2sin^2(x/2)​
    and use the double angle formula to convert cos(x) to a form having sin(x/2). In other words, think of cos(x) as cos(2(x/2)) .

    Then you will have an expression involving only sin(x/2) .
     
  7. Dec 9, 2012 #6
    So basically change cosx into 1-sin^2(x/2) ??
     
  8. Dec 9, 2012 #7

    HallsofIvy

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    NO, cos x is NOT equal to 1- sin^2(x/2). It is equal to 1- sin^2(x). You could then use the identity sin(x)= 2sin(x/2)cos(x/2).
     
  9. Dec 9, 2012 #8
    Sorry for the title too
    Sorry i do not follow why you have the 2 over in 2/(1-cos^2(x/2))
     
  10. Dec 9, 2012 #9
    Thanks, i wasnt too sure about that part..... But isnt it cos^2x that is suppose to equal 1 - sin^2x?
     
  11. Dec 9, 2012 #10

    ehild

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    Apply the half-angle formula: sin2(x/2)=(1-cosx)/2

    ehild
     
  12. Dec 9, 2012 #11

    Yes, i can solve it with the half angle formula but was wondering if i could solve it without that formula....
     
  13. Dec 9, 2012 #12

    ehild

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    You need to use either the half-angle formula or the double-angle one.

    ehild
     
  14. Dec 9, 2012 #13
    How would i approach it with double angle? I followed some of the steps and came with this

    1/1-cos^2(x/2) = 2/cos2(x/2)

    2-2cos^2(x/2) = cos2(x/2)

    Is this right?
     
  15. Dec 9, 2012 #14

    SammyS

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    Not quite:

    cos(x) = 1 - 2sin2(x/2)
     
  16. Dec 9, 2012 #15
    Oh ya, sorry about that, so from here i have:

    1-2sin^2(x/2) = 2sin^2(x/2)

    If i move the right side over i would get:

    1-4sin^2(x/2) = 0

    sin^2(x/2) = 1/4

    Square root it:

    sin(x/2) = 1/2?

    I think thats correct, right?
     
  17. Dec 9, 2012 #16

    ehild

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    sin(x/2) = ±1/2.

    ehild
     
  18. Dec 9, 2012 #17

    Mark44

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    That 2 shouldn't be there. The line I was correcting had 2/(1 - cos(x)), and I brought that 2 along, not noticing that it was wrong as well.
     
  19. Dec 9, 2012 #18
    Thanks., but the answers i get are pi/6 and 5pi/6 while the answers are pi/3 and 5pi/3

    Anything i have to do?
     
  20. Dec 9, 2012 #19

    SammyS

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    If x/2 = π/6 , then x = π/3 ...
     
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