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

1. Michele_Laino

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

yes, i got down to this: csc(2x)=-1

3. Michele_Laino

that's right!

4. Michele_Laino

now, use this substitution: $\Large \csc \left( {2x} \right) = \frac{1}{{\sin \left( {2x} \right)}}$

5. Michele_Laino

you should get this: $\Large \frac{1}{{\sin \left( {2x} \right)}} = - 1$

6. anonymous

how did you get 1/sin(2x) ?

7. Michele_Laino

it is a standard identity. By definition of csc(\alpha), we have: $\Large \csc \alpha \doteq \frac{1}{{\sin \alpha }}$

8. anonymous

oh, i found it tanks.

9. Michele_Laino

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

and we have: $\Large \sin \left( {2x} \right) = 0$ when $\Large \begin{gathered} 2x = 0 \hfill \\ 2x = \pi \hfill \\ \end{gathered}$

11. Michele_Laino

or, when: $\Large \begin{gathered} x = 0, \hfill \\ x = \pi /2 \hfill \\ \end{gathered}$

12. Michele_Laino

so the acceptable solutions, have to be different from 0 and pi/2

13. Michele_Laino

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

the one i found so far is x=3pi/2

15. Michele_Laino

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

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

which is an acceptable solution

18. Michele_Laino

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

where k is an integer number, namely: $\Large k = 0, \pm 1, \pm 2,...$

20. Michele_Laino

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

ok, one after that would be 3pi/4+2pi?

22. Michele_Laino

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

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

thanks

25. Michele_Laino

:)