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Finding solutions for trig!

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Find all solutions for 6sin2x-5sinx+1=0 over the interval [0, 2pi[


    3. The attempt at a solution

    6sin2x-5sinx+1
    (-3sin+1)(-2sin+1)
    1/3 and 1/2
    On unit circle, 1/2 is equal to pi/6 and 5pi/6... Not sure how to figure out 1/3!

    Any help would be much appreciated!
     
  2. jcsd
  3. Jun 2, 2010 #2

    LCKurtz

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    Actually, 1/2 is not equal to pi/6 or 5pi/6 on the unit circle or anywhere else. Perhaps you mean the values of x where sin(x) = 1/2 are pi/6 and 5pi/6.

    As for 1/3, you will have to give an answer numerically. Calculate arcsin(1/3) in radians on your calculator and express the other angle in terms of it.
     
  4. Jun 2, 2010 #3
    You'll have to forgive me for having no idea what I'm talking about... This entire section of trig was thrown at my class during this last week in preparation for exams. To clarify:

    sin(x)=1/3
    x= .34

    Can I go further than this in expressing it? Thanks again for your nearly instant response.
     
  5. Jun 2, 2010 #4

    LCKurtz

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    Ok, yes. But just like in the other case, don't you get another answer in the second quadrant?
     
  6. Jun 2, 2010 #5
    I have absolutely no idea how to find that answer... Could you explain?
     
  7. Jun 2, 2010 #6

    HallsofIvy

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    In your original post you said "On unit circle, 1/2 is equal to pi/6 and 5pi/6". Now you know better than that! "1/2= 0.5" while pi/6 is about .52. What you meant was that sin(pi/6)= 1/2. LCKurtz was suggesting that you be more careful about saying what you mean!

    On the unit circle, sin(t) is the y coordinate of a point at distance t from (1, 0) measured around the circle. Two points on the same horizontal line will have the say y value. If you draw a horizontal line at y= 1/3, it crosses the line twice, at points that are symmetric with respect to the x-axis. Yes one of the points is a distance .3398, approximately, around the circle from (1, 0), and the other point, because it is symmetrically placed, is that same distance from the point (-1, 0). Since (-1, 0) is the entire half circle, a distance of [itex]\pi[/itex] from (1, 0), that symmetrical point is a distance [itex]\pi- .3398[/itex] from it, just as [itex]5\pi/6= \pi- \pi/6[/itex].
     
    Last edited: Jun 2, 2010
  8. Jun 2, 2010 #7


    So, I learned from simplifying that the y coordinate of the points on the unit circle (which is describing the x axis of my function) are 1/2 and 1/3. The y coordinate of 1/2 is described as 30 degrees and 150 degrees on the unit circle, or pi/6 and 5pi/6. To find the other two, I must find the sin of the distance they are equal to. So,

    sin(x)= 1/3
    x=sin-1(1/3)
    x=.3398

    Then to figure out where the second point is, I need to do:
    sin(pi-.3398)= .3333? That doesn't seem to make sense to me. I wrote the above paragraph explaining how I understand this system to work so you could correct me in any instances where I'm wrong.

    EDIT: or is it really as simple as pi-.3398... as in 2.8?
     
    Last edited: Jun 2, 2010
  9. Jun 2, 2010 #8
    So, finally, 6sin2x-5sinx+1=0 over [0,2pi] when x =
    [.3398, 2.8, pi/4, 5pi/6,]
     
  10. Jun 3, 2010 #9

    HallsofIvy

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    Why not? My calculator says that sin(pi- .3398)= sin(2.802)= .33333...

    Yes, it really is! (Although I wouldn't use four significant digits for .3398 and only two for 2.8. I would say, as I did above, that pi- .3398= 2.802.)
     
  11. Jun 9, 2010 #10
    firstly the above is a quadric equation where D = 1

    so [tex]x = \frac{5 \pm 1}{12}[/tex] which implies that x is x = 1/2 or x = 1/3 and if you look in your Calculus book you will find howto convert to radians !
     
    Last edited: Jun 9, 2010
  12. Jun 9, 2010 #11

    Mark44

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    No, the equation in the OP is quadratic in form, but is NOT a quadratic equation. It is also NOT a quadric (fourth-degree) equation.

    Furthermore, the solutions are sin(x) = 1/2 or sin(x) = 1/3, not x = 1/2 or x = 1/3. This was established early on in this thread.
     
  13. Jun 9, 2010 #12

    Char. Limit

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    This is already seemingly over... but I just want to say one thing... isn't a fourth-degree equation called a quartic? Or is "quadric" also a name?

    I'm just curious, really.
     
  14. Jun 9, 2010 #13

    Mark44

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    You're right - quartic is the right term for a fourth-degree equation. Quadric is also a word that is used most often in the term "quadric surfaces," which include paraboloids, ellipsoids, spheres, and hyperboloids.
     
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