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

1. freckles

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

2. freckles

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

3. anonymous

Yes :3

4. freckles

let me know if you need any further help

5. 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.

6. freckles

sin(x) has period 2pi

7. anonymous

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

8. 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

9. anonymous

Thank you!

10. 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

11. anonymous

Oh, i get it better now! Thanks!

12. mathstudent55

You're welcome.