anonymous
  • anonymous
Solve: e^2sinx=1, for 0≤x≤4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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freckles
  • freckles
recall \[e^0=1 \\ \text{ so solve } 2\sin(x)=0 \text{ on } 0 \le x \le 4\]
freckles
  • freckles
also I assumed the problem was \[e^{2 \sin(x)}=1\]
anonymous
  • anonymous
Yes :3

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freckles
  • freckles
let me know if you need any further help
anonymous
  • anonymous
I looked online, and it said the answer was x=0±2πn,π±2πn But i don't understand where the 2πn came from.
freckles
  • freckles
sin(x) has period 2pi
anonymous
  • anonymous
Oh yea! Haha. sorry. It's been a while...
freckles
  • freckles
so sin(x)=sin(x+2pi)=sin(x+4pi)=sin(x+6pi)=... or sin(x)=sin(x+2*npi) where n is an integer
anonymous
  • anonymous
Thank you!
mathstudent55
  • mathstudent55
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
anonymous
  • anonymous
Oh, i get it better now! Thanks!
mathstudent55
  • mathstudent55
You're welcome.

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