Question

Solve: e^2sinx=1, for 0≤x≤4

Answer 1

recall
\[e^0=1 \\ \text{ so solve } 2\sin(x)=0 \text{ on } 0 \le x \le 4\]

Answer 2

also I assumed the problem was
\[e^{2 \sin(x)}=1\]

Answer 3

Yes :3

Answer 4

let me know if you need any further help

Answer 5

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

Answer 6

sin(x) has period 2pi

Answer 7

Oh yea! Haha. sorry.
It's been a while...

Answer 8

so sin(x)=sin(x+2pi)=sin(x+4pi)=sin(x+6pi)=...
or sin(x)=sin(x+2*npi) where n is an integer

Answer 9

Thank you!

Answer 10

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

Answer 11

Oh, i get it better now! Thanks!

Answer 12

You're welcome.