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anonymous
 one year ago
Solve: e^2sinx=1, for 0≤x≤4
anonymous
 one year ago
Solve: e^2sinx=1, for 0≤x≤4

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1recall \[e^0=1 \\ \text{ so solve } 2\sin(x)=0 \text{ on } 0 \le x \le 4\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1also I assumed the problem was \[e^{2 \sin(x)}=1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1let me know if you need any further help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I looked online, and it said the answer was x=0±2πn,π±2πn But i don't understand where the 2πn came from.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sin(x) has period 2pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yea! Haha. sorry. It's been a while...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so sin(x)=sin(x+2pi)=sin(x+4pi)=sin(x+6pi)=... or sin(x)=sin(x+2*npi) where n is an integer

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0Since \(0 \le x \le 4\), 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, i get it better now! Thanks!
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