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anonymous

  • one year ago

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  1. amistre64
    • one year ago
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    what is our derivative of f wrt.x?

  2. anonymous
    • one year ago
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    2-2cos(2x)

  3. amistre64
    • one year ago
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    + not - factor out the 2 and ... what do you see that we can do then?

  4. amistre64
    • one year ago
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    or set it equal to 0, subtract off the 2, divide by -2 etc ...

  5. anonymous
    • one year ago
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    You get 2(1-cosx)

  6. amistre64
    • one year ago
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    2(1 + cos(2x)) = 0 since 2 is never 0, that only leaves 1+cos(2x) to be zero for this to have any chance of working

  7. amistre64
    • one year ago
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    when does 1+cos(2x) = 0?

  8. anonymous
    • one year ago
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    At 0?

  9. amistre64
    • one year ago
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    no cos(0) = 1, not -1

  10. amistre64
    • one year ago
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    cos(2x) = -1 is what we want for 1 + cos(2x) to be zero

  11. amistre64
    • one year ago
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    cos(pi) = -1, so when does 2x = pi?

  12. amistre64
    • one year ago
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    cos(3pi) = -1 as well, any odd multiple of pi makes cos go to -1 2x = 3pi should fit in the interval if i see it correctly

  13. anonymous
    • one year ago
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    Thats the answer?

  14. amistre64
    • one year ago
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    those are approaches that I would take to find the answer ...

  15. anonymous
    • one year ago
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    Which do you think is the best approach

  16. amistre64
    • one year ago
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    ive only given 1 approach

  17. amistre64
    • one year ago
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    given that our interval is from x=0 to pi 2x gives us a range from 0 to 4pi to play with cos(pi) and cos(3pi) are both equal to -1 so we will have 2 solutions.

  18. amistre64
    • one year ago
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    2x= pi, and 2x = 3pi

  19. amistre64
    • one year ago
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    ** our interval is x=0 to 2pi ...

  20. anonymous
    • one year ago
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    So x=pi/2 and x=3/2pi

  21. amistre64
    • one year ago
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    correct

  22. anonymous
    • one year ago
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    Thank you!!!

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