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bridgetx516x Group Title

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

  • one year ago
  • one year ago

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  1. bridgetx516x Group Title
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    The equation written properly is: \[\tan^{2}xcscx=\tan^{2}x\]

    • one year ago
  2. Jamie133 Group Title
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    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 :)

    • one year ago
  3. bridgetx516x Group Title
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    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

    • one year ago
  4. bridgetx516x Group Title
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    @jamie133 have i done it correct so far?

    • one year ago
  5. Jamie133 Group Title
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    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.

    • one year ago
  6. bridgetx516x Group Title
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    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

    • one year ago
  7. Jamie133 Group Title
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    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!

    • one year ago
  8. bridgetx516x Group Title
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    you are correct! thank you!!

    • one year ago
  9. Jamie133 Group Title
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    No problem :) That was a close one, I hadn't noticed the interval!

    • one year ago
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