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

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

Trigonometry
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we can rewrite your function as follows: \[\Large \begin{gathered} 4\csc \left( {2x} \right) - 1 = 9\csc \left( {2x} \right) + 4 \hfill \\ \hfill \\ 5\csc \left( {2x} \right) = - 5 \hfill \\ \end{gathered} \]
yes, i got down to this: csc(2x)=-1
that's right!

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now, use this substitution: \[\Large \csc \left( {2x} \right) = \frac{1}{{\sin \left( {2x} \right)}}\]
you should get this: \[\Large \frac{1}{{\sin \left( {2x} \right)}} = - 1\]
how did you get 1/sin(2x) ?
it is a standard identity. By definition of csc(\alpha), we have: \[\Large \csc \alpha \doteq \frac{1}{{\sin \alpha }}\]
oh, i found it tanks.
now we have to exclude those particular values for which sin(2x)=0, since we are not allowed to divide by zero
and we have: \[\Large \sin \left( {2x} \right) = 0\] when \[\Large \begin{gathered} 2x = 0 \hfill \\ 2x = \pi \hfill \\ \end{gathered} \]
or, when: \[\Large \begin{gathered} x = 0, \hfill \\ x = \pi /2 \hfill \\ \end{gathered} \]
so the acceptable solutions, have to be different from 0 and pi/2
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
the one i found so far is x=3pi/2
ok! since that value is different from x=0, and it is different from x=pi/2, then it is an acceptable solution
sorry, we have: \[\Large 2x = \frac{{3\pi }}{2}\] so, dividing by 2, we get: \[\Large x = \frac{{3\pi }}{4}\] am I right?
which is an acceptable solution
more precisely, the general solution is: \[\Large 2x = \frac{{3\pi }}{2} + 2k\pi \] or, equivalently: \[\Large x = \frac{{3\pi }}{4} + k\pi \]
where k is an integer number, namely: \[\Large k = 0, \pm 1, \pm 2,...\]
so, anothe acceptable solution is given adding pi, to the preceding solution, namely: \[\Large x = \frac{{3\pi }}{4} + \pi = ...\] please complete my computation
ok, one after that would be 3pi/4+2pi?
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
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
thanks
:)

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