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Trig Question

  1. Mar 11, 2005 #1
    sin3x=1/2

    how would i go about solving this problem? and there should be 6 solutions right, how do i get them? thanks.
     
  2. jcsd
  3. Mar 11, 2005 #2
    The answer is x = pi/18 but I don't know how to really get it. I just thought what sin x = 1/2 (x = pi/6) and since its sin 3x, then its that /3 (pi/18).
     
  4. Mar 11, 2005 #3

    BobG

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    There's six solutions because:

    a) there's two angles between 0 and [tex]2 \pi[/tex] that have a sine of 1/2

    b) you have 3x equals one of those two angles. In other words, for each of the angles that have a sine of 1/2, there's three angles between 0 and [tex]2 \pi[/tex] that, if multiplied by 3, would wind up at the same spot on the unit circle.

    Example: [tex]\frac{13 \pi}{18} * 3 = \frac{13 \pi}{6}[/tex] which is the same spot on the unit circle as [tex]\frac{\pi}{6}[/tex]

    Find each of the angles that have a sine of 1/2

    Divide them by 3.

    Add [tex]2 \pi[/tex] to the original angles, then divide by 3 again.

    Add [tex]4 \pi[/tex] to the original angles, then divide by 3 again.

    All three should lie between 0 and 2 pi.

    If you added [tex]6 \pi[/tex] to the original angles and divided by 3, you would find that your answer was greater than [tex]2 \pi[/tex] (should be obvious, since [tex]6 \pi[/tex] divided by 3 is [tex]2 \pi[/tex]).
     
    Last edited: Mar 11, 2005
  5. Mar 11, 2005 #4

    dextercioby

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    [tex] \sin 3x=\sin \ (x+2x)=\sin x \ (\cos^{2}x-\sin^{2}x)+\cos x \ (2\sin x\cos x)=...=-4\sin^{3}x+3\sin x [/tex]...

    The u have to solve a cubic,assuming:
    [tex] \sin x=u [/tex]

    [tex] -4u^{3}+3u-\frac{1}{2}=0 [/tex]...

    Daniel.
     
    Last edited: Mar 11, 2005
  6. Mar 11, 2005 #5

    BobG

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    Wow! That's an interesting solution. Are you solving the same problem he's asking? (He never did actually state what he was solving for. I just kind of assumed he was trying to solve for x.)

    Edit: But, your solution does give the sine of x. A point worth noting, since most people only memorize the basic angles. If you want the sine of pi/18, you would have to multiply your angle by 3 to get an angle you did know the sine of and then solve for sine x the way dex did.
     
    Last edited: Mar 11, 2005
  7. Mar 11, 2005 #6

    dextercioby

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    Bob,you're not funny.

    Daniel.
     
  8. Mar 12, 2005 #7
    didn't have a chance to try anything yet (thanks so much for the help!), but its solving for x, and therefore all the possible angles
     
  9. Mar 12, 2005 #8
    thanks, i think i got that one. now for cox3x=1/2, this is what i did (is it right?)
    ok, so i put 3x = pi/3, 5pi/3. then divide each by 3 to get
    x=pi/9, 5pi/9
    add 2pi/3 and 4pi/3

    for the solutions i get (Π is pi)
    x=Π/9, 5Π/9, 7Π/9, 11Π/9, 13Π/9, 17Π/9

    are these correct for the 6 solutions?
     
    Last edited: Mar 12, 2005
  10. Mar 12, 2005 #9

    dextercioby

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    That is INCOMPATIBLE with the calcuations that u made...

    Daniel.
     
  11. Mar 12, 2005 #10
    you have 2x this time. using 3x will not help you.
     
  12. Mar 12, 2005 #11
    sorry, it was actually cos3x=1/2. can anyone confirm those solutions?
     
  13. Mar 12, 2005 #12
    [tex]\cos{3x} = \frac{1}{2}[/tex]

    So...
    remember that cos is positive in first and fourth quadrant...

    [tex]
    3x = \frac{\pi}{3}, -\frac{\pi}{3}, \frac{5\pi}{3} ....
    [/tex]

    [tex]
    x = \frac{\pi}{9}, -\frac{\pi}{9}, \frac{5\pi}{9} ....
    [/tex]

    Is there a restriction on the angles?? Because you can just keep on adding two pi onto them and it will be a valid answer.
     
  14. Mar 12, 2005 #13
    0 ≤ x < 2pi
    is the restriction

    (and you keep adding 2pi/3, or the period, right?)
     
  15. Mar 12, 2005 #14
    Oopps, this is supposed to be the answers:

    [tex]
    3x = \frac{\pi}{3}, \frac{5\pi}{3} , \frac{7\pi}{3} , \frac{11\pi}{3} , \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3} ... [/tex]

    Because the its in the 1st quadrant and the 4th quadrant, it is pi/3 and 5pi/3, because it is in a circle, you can just keep adding 2pi (360 degrees) to pi/3 or 5pi/3... but when you divide by 3, it has to be within 0 < x < 2pi

    remember that the cos function has a period of 2pi, so it's a repeating cycle -- think of the cos graphs.

    so... u divide each answer by 3...

    [tex]
    x = \frac{\pi}{9}, \frac{5\pi}{9} , \frac{7\pi}{9} , \frac{11\pi}{9} , \frac{13\pi}{9}, \frac{17\pi}{9}
    [/tex]

    As you can see, I didnt include 19pi/9 as an answer because it is outside the domain.
     
    Last edited by a moderator: Mar 12, 2005
  16. Mar 12, 2005 #15

    Curious3141

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    The thing to remember is that for trig equations, there are potentially an *infinite* number of solutions. This is because trig functions are periodic.

    So you must always specify the range you're looking for solutions in. This is usually [tex]0 \leq x < 2\pi[/tex] but not always. So this must be specified.

    Assuming that's the range we're going to use, this is the simplest way to solve it :

    Think about what angles have sines of half. For sine to be positive, the angle can be in the 1st and 2nd quadrants, so you have [tex]\frac{\pi}{6}[/tex] and [tex]\frac{5\pi}{6}[/tex]. Those are two possible values for 3x, correct ?

    Since you're going to be dividing those by 3 to get x, you should take more values for 3x if you want to get all the values in the required range. Since you know [tex]\sin \theta = \sin{(\theta + 2k\pi)}[/tex], where k is an integer, just add [tex]2\pi[/tex] and [tex]4\pi[/tex] to the above values to get all the possible values of 3x.

    So you have :

    [tex]3x = \frac{\pi}{6}, \frac{5\pi}{6}, (\frac{\pi}{6} + 2\pi), (\frac{5\pi}{6} + 2\pi), (\frac{\pi}{6} + 4\pi), (\frac{5\pi}{6} + 4\pi)[/tex]

    and dividing by 3 and simplifying, you get the required solutions in radians within the required range.
     
  17. Mar 13, 2005 #16
    ok, thanks to everyone, i got the answers (which i probably could've done) but more importantly know how to do it and why it works. thanks!
     
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