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Trig Equation

  1. Feb 27, 2004 #1
    For some reason, I seem to be unable to algebraically solve this equation:

    sin(x) + sqrt(3)cos(x) = 1

    Any help would be appreciated.
     
  2. jcsd
  3. Feb 27, 2004 #2

    Hurkyl

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    You need to combnie the LHS into a single trig function.
     
  4. Feb 27, 2004 #3
    Never mind, LHS means left hand side.

    Yes, I know I need to convert the left side into the same trig function. That is what I'm having trouble with.
     
  5. Feb 27, 2004 #4

    Hurkyl

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    All righty.

    Suppose the equation was of the form:

    [tex]
    \cos \frac{\pi}{5} \sin x + \sin \frac{\pi}{5} \cos x = 1
    [/tex]

    Would you be able to solve for x?
     
  6. Feb 27, 2004 #5
    Yes, but I don't know how exactly that is applied here.
     
  7. Feb 27, 2004 #6

    Hurkyl

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    (I should've mentioned that there will be a couple steps to this)


    Ok. pretend for a moment that you could solve the equations:

    cos y = 1
    sin y = √3

    Then would you be able to solve the equation:

    sin x + √3 cos x = 1
     
  8. Feb 27, 2004 #7

    matt grime

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    There is a general formula for this, usuallr referred to as rsin(theta + x)

    but here, have you thought about multiplying everything by the same number so you get something akin to Hurkyl's example (think of some obvious values of cos sin etc involving sqrt(3))?
     
  9. Feb 27, 2004 #8
    I'm sorry, but still nope :(
     
  10. Feb 27, 2004 #9

    Hurkyl

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    So you know how to solve the equation:

    cos y sin x + sin y cos x = z

    for x, if you know what y and z are.


    Now, if I want to solve the equation

    A sin x + B cos x = z

    and I know that

    A = cos y
    and
    B = sin y

    Then can you solve this equation for x?
     
  11. Feb 27, 2004 #10
    Hmm, I think I figured it out. Tell me if I am right:

    cos(y) = z
    sin(y) = z*sqrt(3)

    So y = tan^-1(sqrt(3)) = pi/3

    So

    sin(x)cos(y) - cos(x)sin(y) = 1z
    sin(x-y) = 1z
    x-y = sin^-1(.5)

    and then solve for x?

    Thanks a lot
     
  12. Feb 27, 2004 #11
    is there any significance to the value z (in my previous post) always seeming to equal 1/sqrt(A^2 + B^2)?
     
  13. Feb 27, 2004 #12

    Hurkyl

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    Well, what does [itex]\sin^2 x + \cos^2 x[/itex] equal?
     
  14. Feb 27, 2004 #13
    That makes sense! I can't beleive I didn't figure this problem out myself.

    Thanks a lot for the help.
     
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