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anonymous

  • one year ago

Solve: e^2sinx=1, for 0≤x≤4

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  1. freckles
    • one year ago
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    recall \[e^0=1 \\ \text{ so solve } 2\sin(x)=0 \text{ on } 0 \le x \le 4\]

  2. freckles
    • one year ago
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    also I assumed the problem was \[e^{2 \sin(x)}=1\]

  3. anonymous
    • one year ago
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    Yes :3

  4. freckles
    • one year ago
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    let me know if you need any further help

  5. anonymous
    • one year ago
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    I looked online, and it said the answer was x=0±2πn,π±2πn But i don't understand where the 2πn came from.

  6. freckles
    • one year ago
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    sin(x) has period 2pi

  7. anonymous
    • one year ago
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    Oh yea! Haha. sorry. It's been a while...

  8. freckles
    • one year ago
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    so sin(x)=sin(x+2pi)=sin(x+4pi)=sin(x+6pi)=... or sin(x)=sin(x+2*npi) where n is an integer

  9. anonymous
    • one year ago
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    Thank you!

  10. mathstudent55
    • one year ago
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    Since \(0 \le x \le 4\), doesn't that restrict the solutions to only a few values of x, and not x=0±2πn,π±2πn in general for any integer n. The solutions are only x = 0, π

  11. anonymous
    • one year ago
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    Oh, i get it better now! Thanks!

  12. mathstudent55
    • one year ago
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    You're welcome.

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