A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
how would you find the solution for this
sin^2x  .49=0< the answer has 4 different solutions
anonymous
 4 years ago
how would you find the solution for this sin^2x  .49=0< the answer has 4 different solutions

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(sinx)^2 =.49 sin(x) = +/ .49 doesn't give an interval, so it has infinite solutions... if your interval is something like 0..2pi then arcsin(.49) (two solutions) and arcsin(.49) (two more solutions)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the intervals are between 0 and 2pie

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and wouldn't it be root .49 because sin is squrd?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(sinx)^2 =.49 sin(x) = +/ sqrt(.49) doesn't give an interval, so it has infinite solutions... if your interval is something like 0..2pi then arcsin(sqrt(.49) (two solutions) and arcsin(sqrt(.49) ) (two more solutions)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but how would you find the solution would i add 3.14 to root .49 and subtract 3.14 (btw everythings in radians)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.