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bridgetx516x
 one year ago
Best ResponseYou've already chosen the best response.0The equation written properly is: \[\tan^{2}xcscx=\tan^{2}x\]

Jamie133
 one year ago
Best ResponseYou've already chosen the best response.1Try 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 :)

bridgetx516x
 one year 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

bridgetx516x
 one year ago
Best ResponseYou've already chosen the best response.0@jamie133 have i done it correct so far?

Jamie133
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

bridgetx516x
 one year 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

Jamie133
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm, 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!

bridgetx516x
 one year ago
Best ResponseYou've already chosen the best response.0you are correct! thank you!!

Jamie133
 one year ago
Best ResponseYou've already chosen the best response.1No problem :) That was a close one, I hadn't noticed the interval!
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