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Radians/Trigonometric Equations

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve for x, where [tex]0\leq x < 2\pi[/tex]. Then give a general solution.
    The question:
    [tex]sin\²x-1=0[/tex]

    2. Relevant equations
    Perhaps [tex]sin\theta=\frac{y}{r}[/tex]?
    None else that I'm aware of.

    3. The attempt at a solution
    So far I obviously have [tex]sin\²x=1[/tex]. Using a graphing calculator I have solved the problem, however I am unsure how to solve this algebraically. With that, I leave you with my question.
     
  2. jcsd
  3. Jul 19, 2008 #2
    where is [tex]\frac{y}{r}=1?[/tex]
    I'm looking at the unit circle....
     
  4. Jul 19, 2008 #3
    Well I just mean [tex]\frac{y}{r}[/tex] is [tex]\frac{opposite}{hypotenuse}[/tex]. What is the unit circle?

    In general I'm just finding it difficult to do this type of question as the textbook is not explaining it very clearly. The answer in the back is:

    [tex]\frac{\pi}{2},\frac{3\pi}{2};(2n+1)\frac{\pi}{2},[/tex] n any integer
     
    Last edited: Jul 19, 2008
  5. Jul 19, 2008 #4

    Dick

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    Look at a graph of sin. It's periodic with period 2*pi meaning that the value of sin(x) and sin(x+2*pi) are equal. The only solution in [0,2*pi] is pi/2. The other solutions are just periodic extensions of that. And 3*pi/2 isn't one of them. What's up with that?
     
    Last edited: Jul 19, 2008
  6. Jul 20, 2008 #5
    What do you mean by the solution for [tex][0,2\pi][/tex]? I kind of understand where you explained that [tex]\frac{\pi}{2}[/tex] was the solution of where sinx=1, but I'm not sure like you said how is [tex]\frac{3\pi}{2}[/tex] an answer? [tex]\frac{\pi}{2}[/tex] is equivalent to 90° which satisfies [tex]sin\theta=1[/tex], but [tex]\frac{3\pi}{2}[/tex] is equivalent to 270° which would be -90°, which wouldn't be a solution right?

    PS I'm sure I might be repeating a lot of what you said, I'm just trying to make sure that I understand correctly.
     
  7. Jul 20, 2008 #6

    Dick

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    I was just pointing out that the answer in the back is not correct. E.g. 3*pi/2 is WRONG.
     
  8. Jul 20, 2008 #7
    So would that mean that [tex](2n+1)\frac{\pi}{2}[/tex] is wrong? Because if:

    Let n be 1
    [tex](2(1)+1)\frac{\pi}{2}[/tex]
    [tex](2+1)\frac{\pi}{2}[/tex]
    [tex]\frac{3\pi}{2}[/tex]

    Then it appears to be correct
     
  9. Jul 20, 2008 #8

    arildno

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    Are you sure the equation given in the book wasn't:
    [tex]\sin^{2}(x)-1=0[/tex]
    In that case, the answers in your book is correct.
     
  10. Jul 20, 2008 #9
    Oh wow you're exactly right, I'm sorry about that, I guess I can't re-edit my original post.

    How would I go about solving this?
     
  11. Jul 20, 2008 #10

    arildno

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    Well, that equation implies that:
    [tex]\sin(x)=\pm{1}[/tex]
    Thus, you must combine the solutions of the two equations:
    [tex]\sin(x)=1[/tex] and [tex]\sin(x)=-1[/tex] in order to get the full solution set.
     
  12. Jul 20, 2008 #11
    Oh ok so that's why there would be 90° and -90°?
     
  13. Jul 20, 2008 #12

    arildno

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    That's correct! :smile:
     
  14. Jul 20, 2008 #13
    Ok thanks, and I'm really sorry that I screwed up on that, I'll be sure to check harder next time. Thanks.
     
  15. Jul 20, 2008 #14
    Oh one more question, I don't understand how (2n+1) becomes part of the answer, can someone explain? Thank you.
     
  16. Jul 20, 2008 #15

    Dick

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    The x values where sin(x)^2=1 are pi/2, 3*pi/2, 5*pi/2, 7*pi/2 etc etc. (2n+1) is a concise way of expressing the numerator of those values.
     
  17. Jul 21, 2008 #16
    Ok I see, and I figured I'd ask this instead of making a new thread. When trying to graph the functions on a graphing calculator (I use the ti-83+), how do I know what to make the window size (X scale/Y scale)? Because in my textbook it mentions things like "Since solutions are to be in the domain [tex]0\leq x<2\pi[/tex], choose radian mode and use a window with Xmin = 0, Xmax = [tex]2\pi[/tex], and Xscl (X Scale)=[tex]\frac{\pi}{2}[/tex]. I just don't understand how they know what to set the x scale to so accurately. Thanks.
     
  18. Jul 21, 2008 #17
    Ok nevermind I figured it out on my own, sorry for this double post, I'm done with this thread.
     
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