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anonymous

  • one year ago

4 csc(2x)-1=9 csc(2x)+4 specified interval [0,2pi]

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  1. Michele_Laino
    • one year ago
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    we can rewrite your function as follows: \[\Large \begin{gathered} 4\csc \left( {2x} \right) - 1 = 9\csc \left( {2x} \right) + 4 \hfill \\ \hfill \\ 5\csc \left( {2x} \right) = - 5 \hfill \\ \end{gathered} \]

  2. anonymous
    • one year ago
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    yes, i got down to this: csc(2x)=-1

  3. Michele_Laino
    • one year ago
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    that's right!

  4. Michele_Laino
    • one year ago
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    now, use this substitution: \[\Large \csc \left( {2x} \right) = \frac{1}{{\sin \left( {2x} \right)}}\]

  5. Michele_Laino
    • one year ago
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    you should get this: \[\Large \frac{1}{{\sin \left( {2x} \right)}} = - 1\]

  6. anonymous
    • one year ago
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    how did you get 1/sin(2x) ?

  7. Michele_Laino
    • one year ago
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    it is a standard identity. By definition of csc(\alpha), we have: \[\Large \csc \alpha \doteq \frac{1}{{\sin \alpha }}\]

  8. anonymous
    • one year ago
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    oh, i found it tanks.

  9. Michele_Laino
    • one year ago
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    now we have to exclude those particular values for which sin(2x)=0, since we are not allowed to divide by zero

  10. Michele_Laino
    • one year ago
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    and we have: \[\Large \sin \left( {2x} \right) = 0\] when \[\Large \begin{gathered} 2x = 0 \hfill \\ 2x = \pi \hfill \\ \end{gathered} \]

  11. Michele_Laino
    • one year ago
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    or, when: \[\Large \begin{gathered} x = 0, \hfill \\ x = \pi /2 \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    so the acceptable solutions, have to be different from 0 and pi/2

  13. Michele_Laino
    • one year ago
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    now our last equation is equivalent to this one: \[\Large \sin \left( {2x} \right) = - 1\] please solve it, for x, what do you get? @quocski

  14. anonymous
    • one year ago
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    the one i found so far is x=3pi/2

  15. Michele_Laino
    • one year ago
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    ok! since that value is different from x=0, and it is different from x=pi/2, then it is an acceptable solution

  16. Michele_Laino
    • one year ago
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    sorry, we have: \[\Large 2x = \frac{{3\pi }}{2}\] so, dividing by 2, we get: \[\Large x = \frac{{3\pi }}{4}\] am I right?

  17. Michele_Laino
    • one year ago
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    which is an acceptable solution

  18. Michele_Laino
    • one year ago
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    more precisely, the general solution is: \[\Large 2x = \frac{{3\pi }}{2} + 2k\pi \] or, equivalently: \[\Large x = \frac{{3\pi }}{4} + k\pi \]

  19. Michele_Laino
    • one year ago
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    where k is an integer number, namely: \[\Large k = 0, \pm 1, \pm 2,...\]

  20. Michele_Laino
    • one year ago
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    so, anothe acceptable solution is given adding pi, to the preceding solution, namely: \[\Large x = \frac{{3\pi }}{4} + \pi = ...\] please complete my computation

  21. anonymous
    • one year ago
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    ok, one after that would be 3pi/4+2pi?

  22. Michele_Laino
    • one year ago
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    yes! nevertheless it is greater than 2*pi, and your problem asks for all solutions such that are greater or equal to zero and are less or equal to 2*pi

  23. Michele_Laino
    • one year ago
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    so we have these 2 solutions: \[\Large \begin{gathered} x = \frac{{3\pi }}{4}, \hfill \\ \\ x = \frac{{7\pi }}{4} \hfill \\ \end{gathered} \] which are both acceptable, since they are different from 0 and from pi/2

  24. anonymous
    • one year ago
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    thanks

  25. Michele_Laino
    • one year ago
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    :)

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