A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DominiRican1013

  • one year ago

Solve on the interval [0,2π): 2cos2x+3cosx+1=0

  • This Question is Closed
  1. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    here we have to make this substitution: \[\cos \left( {2x} \right) = 2{\left( {\cos x} \right)^2} - 1\]

  2. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so your equation, can be rewritten as follows: \[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) + 3\cos x + 1 = 0\]

  3. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please simplify

  4. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what equation do you get?

  5. DominiRican1013
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know I'm trying to work it out

  6. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hint: \[4{\left( {\cos x} \right)^2} - 2 + 3\cos x + 1 = 0\]

  7. DominiRican1013
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll be right back.

  8. DominiRican1013
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know how to do this.

  9. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    more explanation: we have: \[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) = 4{\left( {\cos x} \right)^2} - 2\] am I right?

  10. DominiRican1013
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    these are the choices \[A. x=2\pi, x=\pi/3\] \[B. x=\pi, x=2\pi/3, x=4\pi/3\] \[C. x=2\pi, x=\pi/4, x=5\pi/4\] \[D. x=\pi/6, x=7\pi/6\]

  11. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes! I see them, nevertheless I can not give you the answer directly, since I have to respect the Code of Conduct

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If I simplify my equation above, I get: \[4{\left( {\cos x} \right)^2} + 3\cos x - 1 = 0\] now, I make this substitution: cos(x)=y, so I can write: \[4{y^2} + 3y - 1 = 0\] which is a quadratic equation. please solve that equation for y

  13. DominiRican1013
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    B is the answer I got it

  14. 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

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.