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anonymous
 one year ago
4 csc(2x)1=9 csc(2x)+4 specified interval [0,2pi]
anonymous
 one year ago
4 csc(2x)1=9 csc(2x)+4 specified interval [0,2pi]

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we can rewrite your function as follows: \[\Large \begin{gathered} 4\csc \left( {2x} \right)  1 = 9\csc \left( {2x} \right) + 4 \hfill \\ \hfill \\ 5\csc \left( {2x} \right) =  5 \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, i got down to this: csc(2x)=1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, use this substitution: \[\Large \csc \left( {2x} \right) = \frac{1}{{\sin \left( {2x} \right)}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3you should get this: \[\Large \frac{1}{{\sin \left( {2x} \right)}} =  1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get 1/sin(2x) ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is a standard identity. By definition of csc(\alpha), we have: \[\Large \csc \alpha \doteq \frac{1}{{\sin \alpha }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, i found it tanks.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now we have to exclude those particular values for which sin(2x)=0, since we are not allowed to divide by zero

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3and we have: \[\Large \sin \left( {2x} \right) = 0\] when \[\Large \begin{gathered} 2x = 0 \hfill \\ 2x = \pi \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3or, when: \[\Large \begin{gathered} x = 0, \hfill \\ x = \pi /2 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so the acceptable solutions, have to be different from 0 and pi/2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now our last equation is equivalent to this one: \[\Large \sin \left( {2x} \right) =  1\] please solve it, for x, what do you get? @quocski

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the one i found so far is x=3pi/2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! since that value is different from x=0, and it is different from x=pi/2, then it is an acceptable solution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3sorry, we have: \[\Large 2x = \frac{{3\pi }}{2}\] so, dividing by 2, we get: \[\Large x = \frac{{3\pi }}{4}\] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3which is an acceptable solution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3more precisely, the general solution is: \[\Large 2x = \frac{{3\pi }}{2} + 2k\pi \] or, equivalently: \[\Large x = \frac{{3\pi }}{4} + k\pi \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3where k is an integer number, namely: \[\Large k = 0, \pm 1, \pm 2,...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so, anothe acceptable solution is given adding pi, to the preceding solution, namely: \[\Large x = \frac{{3\pi }}{4} + \pi = ...\] please complete my computation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, one after that would be 3pi/4+2pi?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes! 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so 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
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