Solve: e^2sinx=1, for 0≤x≤4

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Solve: e^2sinx=1, for 0≤x≤4

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

recall \[e^0=1 \\ \text{ so solve } 2\sin(x)=0 \text{ on } 0 \le x \le 4\]
also I assumed the problem was \[e^{2 \sin(x)}=1\]
Yes :3

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question