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Trig Equation

  1. Sep 6, 2006 #1
    Find a solution to the equation if possible.
    Give the answer in exact form and in decimal form.

    1 = 8cos(2x + 1) - 3
    1 = 8tan(2x + 1) - 3

    I don't know how to do this one.. but I know how to do the simpler ones like..

    2 = 5sin(3x)
    2/5 = sin(3x)
    (sin-1(2/5)) / 3 = 0.137172

    Those are easier... but the one i've said before...

    1 = 8cos(2x + 1) - 3 is tough..
    1/8 = cos(2x + 1) - 3
    3.125 = cos(2x + 1)
    {[cos-1(3.125)] / 2} - 1

    Is this correct thus far, and if it is... cos-1(3.125) is nonreal. So I guess there are no solutions...

    How about this one?

    1 = 8tan(2x + 1) - 3
    1/8 = tan(2x + 1) - 3
    3.125 = tan(2x + 1)
    {[tan-1(3.125)] / 2} - 1 = -0.369453

    But, the right side doesn't evaluate to 1 when you plug -0.369453 in for x :frown:

    I don't know where to go with these.
     
  2. jcsd
  3. Sep 6, 2006 #2
    First, this is a "calc and beyond" and this is a trig question. Second, call the argument something like u. The problem "looks" a whole lot easier this way. Next try chaning the tangent function into another form, namely sines and cosines. You'll find the answer falls out rather quickly.
     
  4. Sep 7, 2006 #3

    TD

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    Homework Helper

    Ouch! You probably learnt that multiplying by 8 on one side, is dividing by it on the other side.
    However: there's the -3 that has to go first! It's better to remember the rules:
    - you may multiply both sides with a non-zero number
    - you may add the same number to both sides

    Now, if you'd want to get rid of the 8, you'd have to multiply both sides with 1/8, but that would also give -3/8 on the right side, and no longer a -3.
    What you would probably do is first add 3 to both sides (i.e. changing sides of -3), giving 1+3 = 8cos(2x + 1). Now you can remove the 8, giving:

    4/8 = cos(2x + 1) <=> cos(2x + 1) = 1/2.

    And a cosine can surely be 1/2, so this will have a solution, actually two.
    You're making the same mistake with the tan-equation.
     
    Last edited: Sep 7, 2006
  5. Sep 7, 2006 #4
    haha, sometimes it's fun to laugh at our own stupidity...

    Why did I think the right side was a monomial for some reason? I guess I just didn't look at it like I should have. This is a lot easier than I was making it out to be.

    Thanks for pointing out my idiocy though lol.
     
  6. Sep 7, 2006 #5
    so, let's try that again...

    1 = 8cos(2x + 1) - 3
    4 = 8cos(2x + 1)
    1/2 = cos(2x + 1)

    Where do I go from here? .. I'm not sure

    {[cos-1(1/2)] / 2} - 1 = -0.476401

    But that answer doesn't work when plugged back in for x..
     
  7. Sep 7, 2006 #6
    You made the same mistake again, you got to cos-1(1/2)=2x+1 OK, but then you divided by 2 without dividing the 1!
     
  8. Sep 7, 2006 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    cos(u)= 1/2 gives [itex]u= \frac{2\pi}{3}[/itex] as principal value.
    Now, you have [itex]2x+1= \frac{2\pi}{3}[/itex]. Solve that for x.
     
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