Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Atkinsoha

  • 3 years ago

PreCalc & Trig, picture below, please show your work.. Really frustrated with this homework. :/

  • This Question is Closed
  1. Atkinsoha
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. ghass1978
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use sinx=2sin(x/2)cos(x/2) then factorize

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

    Maybe I should tell you this: I haven't been in class for over a week. I don't know any of this stuff.. so be as simple and thorough as you can, thanks.

  4. getusel
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\cos(\frac{ x }{ 2 })-sinx=0\] use the identity sin2x=2sinxcosx \[\cos(x/2)-2\sin(x/2)\cos(x/2)=0\] \[\cos(x/2)(1-\sin(x/2))=0\] cos(x/2)=0 or 1-sin(x/2)=0 \[\cos(x/2)=0 => x/2=\pi/2 \] \[x=\pi\] \[1-\sin(\frac{ x }{ 2 })=0\]

  5. getusel
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sin(\frac{ x }{ 2 })=1 => \frac{ x }{ 2 }=\pi/2\] \[x=\pi\]

  6. getusel
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the solution is \[\pi\]

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

    okay, thank you.

  8. 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