Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

dragonsly

  • 3 years ago

how would you find the solution for this sin^2x - .49=0<--- the answer has 4 different solutions

  • This Question is Closed
  1. Algebraic!
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  2. dragonsly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the intervals are between 0 and 2pie

  3. dragonsly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and wouldn't it be root .49 because sin is squrd?

  4. Algebraic!
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good call...

  5. Algebraic!
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  6. dragonsly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but how would you find the solution would i add 3.14 to root .49 and subtract 3.14 (btw everythings in radians)

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy