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anonymous
 3 years ago
Solve the equation tan^2xcscx=tan^2x on [0, 2π ).
anonymous
 3 years ago
Solve the equation tan^2xcscx=tan^2x on [0, 2π ).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The equation written properly is: \[\tan^{2}xcscx=\tan^{2}x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Try rearranging the equation by bringing everything over to the left. Then factorise out tan^2(x). Then you will have a product. Equate both bits equal to zero and solve them. :) I'm not sure that that is very clear, so let me know :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well what I did was: tan^2xcscxtan^2x=0 and then i tried to factor out tanx tanx(tanxcscxtanx) then i realized i should factor out tan^2x tan^2x (cscx1) and i made those 2 seperate equations: tan^2x=0 and cscx=1 with the cscx=1 i got pi/2 however i am unsure what to do with the tan^2x=0 because of the square

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@jamie133 have i done it correct so far?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think pi/2 is correct! And to get rid of the square, square root both sides. You can do this without worrying about getting a plus or minus, because it is equal to 0. There is no 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it would be tanx=0 giving me 0 and pi. i am not sure if this is correct because it is a multiple choice problem and the answers are: A. x=0 B. x=0 or x=pi C. x=pi/2 or 3pi/2 D. no solution if i get 0, pi and pi/2 which would be the correct answer @jamie133

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm, well it seems that you're working on a funny interval. Because on [0,2pi), tan(x) isn't always defined, and neither is cosec(x). If you actually substitute in 0, pi or pi/2, then you should find that for each one, either tan(x) or cosec(x) is undefined. So, I'm thinking that there are not any solutions for this!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you are correct! thank you!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No problem :) That was a close one, I hadn't noticed the interval!
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