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Trig: general solution

  1. Jun 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Given: sec[tex]\theta[/tex]=[tex]\sqrt{10}[/tex] where 0< [tex]\theta[/tex] <90

    and [tex]\sqrt{10}[/tex]sin(A-[tex]\theta[/tex])=sinA-3cosA



    Determine the solution of
    6cosA +3 = 2sinA


    for A [tex]\in[/tex] [-180; 180], rounded off to one decimal digit.

    2. Relevant equations



    3. The attempt at a solution

    3=2sinA - 6cosA

    3=2(sinA-3cosA)

    [tex]\frac{3}{2}[/tex]=sinA-3cosA

    [tex]\frac{3}{2}[/tex]=[tex]\sqrt{10}[/tex]sin(A-[tex]\theta[/tex])

    Now I can't get rid of the theta
     
  2. jcsd
  3. Jun 15, 2009 #2

    rock.freak667

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    if [itex]sec \theta = \sqrt{10}[/itex] that means [itex]cos \theta = ?[/itex]
     
  4. Jun 15, 2009 #3
    cos[tex]\theta[/tex]=1/[tex]\sqrt{10}[/tex]
    but that still doen't help with getting rid of the theta over here.

    3/2=[tex]\sqrt{10}[/tex]sin(A-[tex]\theta[/tex])
     
  5. Jun 15, 2009 #4

    rock.freak667

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    erm...


    [tex]cos \theta = \frac{1}{\sqrt{10}}[/tex]

    [itex]\theta[/itex] is what then in the range [itex]0< \theta < 90[/itex] ?
     
  6. Jun 15, 2009 #5

    got it Thank you.
     
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