## bridgetx516x 2 years ago Solve the equation tan^2xcscx=tan^2x on [0, 2π ).

1. bridgetx516x

The equation written properly is: \[\tan^{2}xcscx=\tan^{2}x\]

2. Jamie133

Try 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 :)

3. bridgetx516x

well what I did was: tan^2xcscx-tan^2x=0 and then i tried to factor out tanx tanx(tanxcscx-tanx) then i realized i should factor out tan^2x tan^2x (cscx-1) 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

4. bridgetx516x

@jamie133 have i done it correct so far?

5. Jamie133

I 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.

6. bridgetx516x

so 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

7. Jamie133

Hmmm, 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!

8. bridgetx516x

you are correct! thank you!!

9. Jamie133

No problem :) That was a close one, I hadn't noticed the interval!