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PLEASE HELP, need this today

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve 2sinx+1=tanx 0[tex]\leq[/tex]x[tex]\prec[/tex]2 pi

    2. Relevant equations



    3. The attempt at a solution
    I tried to put it in this way
    2sinx+1=tanx
    2sinx+1=(sinx/cosx)
    2sinxcosx+cosx-sinx=0
    sin2x+cosx-sinx=0
    I don't know what to do next, I am stuck, plz help!
    Thanks

    Btw , can anyone show me the step for this one
    2cotxsinx+cotx-2sinx-1=0 domain:0[tex]\leq[/tex]x[tex]\prec[/tex]2 pi
    thanks!
     
  2. jcsd
  3. Nov 23, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    What is sin2x-sinx equal to?

    (Hint: Sum to product formula)
     
  4. Nov 24, 2009 #3
    I haven't learn that yet...
    Sin(a+b),cos(a+b)...
    Sin2A=2sinAcosA, cos2A=cosx^2-sin^2....
    and sinx^2+cosx^2=1 ....and other form
    this is all i know
    We just finished this unit today, i don't think i will learn that...

    Can i square both side?
    sin2x+cosx-sinx=0
    (sin2x)^2=(sinx-cosx)^2
    (sin2x)^2=sinx^2-2cosxsinx+cosx^2
    (sin2x)^2=1-sin2x
    sin2x^x+sin2x-1=0
    and then use quadratic
     
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4

    rock.freak667

    User Avatar
    Homework Helper

    yes that will work as well
     
  6. Nov 25, 2009 #5
    http://www.wolframalpha.com/input/?i=2sinx+1-tanx=0
    http://www.wolframalpha.com/input/?i=sin2x^2%2Bsin2x-1%3D0
    But it seem that both graph is different...
    where did i go wrong??
     
  7. Nov 25, 2009 #6

    Mark44

    Staff: Mentor

    What did you get when you solved this equation?
    sin2x^x+sin2x-1=0
     
  8. Nov 25, 2009 #7
    i got 4 solution,
    but 2sinx+1=tanx only have 2 solution..
     
  9. Nov 25, 2009 #8

    Mark44

    Staff: Mentor

    Squaring both sides of an equation often introduces extraneous solutions that aren't solutions of the original, unsquared equation. Check each of your four solutions in your original equation. Keep only the ones that satisfy it.
     
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