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anonymous

  • one year ago

Thanks for helping! I am confused on how many solutions I can get for sin(2x)=sqrt(3)/2. I know that it goes on and I can +2*pi*k to the end. But should I add it to just pi/6 as my answer? Or should I also write pi/3+2*pi*k along with pi/6+2*pi*k? Thanks!

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  1. anonymous
    • one year ago
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    @Kainui @triciaal @jim_thompson5910 @ganeshie8 @Loser66 @robtobey

  2. anonymous
    • one year ago
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    @zepdrix @sammixboo @pooja195 @radar

  3. anonymous
    • one year ago
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    @Whitemonsterbunny17 @triciaal @sleepyjess

  4. anonymous
    • one year ago
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    It wouldn't be 2kpi in this case actually. So consider this: If we were to solve this equation as if it were sin(x) = sqrt(3)/2, we would have these solutions: \(x = \frac{\pi}{3} + 2k\pi\) and \(x = \frac{2\pi}{3} + 2k\pi\) But our actually angle is 2x, so I want to substitute 2x for x in the solutions I have: \(2x = \frac{\pi}{3} + 2k\pi\) and \(2x = \frac{2\pi}{3} + 2k\pi\) If you were to solve for x you would have: \(x = \frac{\pi}{6} + k\pi\) and \(x = \frac{\pi}{3} + k\pi\) Notice how having an angle that is not x causes the periodic behavior of the solutions to change as well.

  5. anonymous
    • one year ago
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    Oh gotcha, so the 2 in the period gets cancelled from the 2x

  6. anonymous
    • one year ago
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    Yes. So the solutions you have, including the +2kpi part, is divided by the angle you have. If it were sin(3x), it'd become + 2kpi/3, etc.

  7. anonymous
    • one year ago
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    I know that pi/3 is one of the final answers, but am somewhat confused on how to get it. I know that the other one is found by arcsin. Is it okay if you show me how to get pi/3? Thanks!

  8. anonymous
    • one year ago
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    @Concentrationalizing

  9. anonymous
    • one year ago
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    Well, pi/3 is because sinx has a solution at 2pi/3. But if you divide by the 2 in the angle 2x, it reduces to pi/3.

  10. anonymous
    • one year ago
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    I know that part, but how to get 2pi/3

  11. anonymous
    • one year ago
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    it is covered by the +pi*k right? So I don't need to include it?

  12. anonymous
    • one year ago
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    Well, I always start by solving the equation as if the angle were simply x. Thats why I started by solving the equation sinx = sqrt(3)/2. Me solving that equation is where I get 2pi/3 from. Now, im not saying 2pi/3 is a solution to the actual problem, just that it is a solution to sinx = 2pi/3. But getting all the solutions to sinx = sqrt(3)/2 well then let me get all solutions to sin(ax) where a could be any number. The solutions you have, the pi/3 and pi/6 are correct and they come from dividing the solutions of sinx = sqrt(3)/2 by 2. All you need to write as a final answer though is pi/3 + kpi, pi/6 + kpi.

  13. anonymous
    • one year ago
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    Okay thanks, that was really informative! I gave you a medal!

  14. anonymous
    • one year ago
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    You're welcome. Good luck! :)

  15. anonymous
    • one year ago
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    :)

  16. phi
    • one year ago
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    ***I know that pi/3 is one of the final answers, but am somewhat confused on how to get it. I know that the other one is found by arcsin. **** assume we are solving \[ 2x = \sin^{-1}\left( \frac{\sqrt{3}}{2}\right) \approx 0.866 \] you can look at the graph of sin y |dw:1433703018354:dw| and can see that there are two places where the sin y = 0.866 one of them is given by your calculator using inverse sin (you get 60 degrees or pi/3) (and after you divide by 2, you will get x = pi/6+ k pi ) the other solution is "on the other side" of 90 degrees the same distance away as the first solution. in other words, the answer 60 is 30 to the left of 90, so the other solution is 30 to the right of 90... which is 120 degrees or 2pi/3 after dividing by 2 you get x= pi/3 + k pi if you "figure out" the answer for a few different "k" values you see they are different numbers and you need both x = pi/6+ k pi x= pi/3 + k pi

  17. anonymous
    • one year ago
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    Thanks phi!

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