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helpplease1 Group Title

Determine the general formula(s) for the solution. Then, determine the specific solutions on the interval [0,2pi). 2cos^2(theta)-3cos(theta)+1=0

  • one year ago
  • one year ago

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  1. looser66 Group Title
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    still need help?

    • one year ago
  2. helpplease1 Group Title
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    yes i do :(

    • one year ago
  3. looser66 Group Title
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    let cos\(\theta\)= x, you have the form of \(2x^2-3x+1=0\) solve for x as a quadratic equation, then plulg back to x = cos \(\theta\) to solve for \(\theta\)

    • one year ago
  4. helpplease1 Group Title
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    thank you! but i mistyped the question it is actually plus 3, not 1. :( so would the quadratic equation be \[2x ^{2}-3x+3\]?

    • one year ago
  5. helpplease1 Group Title
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    i cannot figure out how to factor that

    • one year ago
  6. looser66 Group Title
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    which class are you in? do you know Euler's formula for complex number?

    • one year ago
  7. helpplease1 Group Title
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    i am just in a college trig class. 1316. we have not gotten to euler's formula thus far

    • one year ago
  8. helpplease1 Group Title
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    this is the actual problem \[2\sin ^{2}\Theta-3\cos \Theta=-3\]

    • one year ago
  9. looser66 Group Title
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    hey, it totally different,!! heehe , let see. 2 sin^2 = 2 - 2cos^2 therefore, 2-2cos^2 -3 cos +3 =0 2cos^2 +3 cos -5 =0, can solve it now

    • one year ago
  10. looser66 Group Title
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    you should post the original one, friend.

    • one year ago
  11. looser66 Group Title
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    you really scared me!! I thought I met a Ph.D student.

    • one year ago
  12. helpplease1 Group Title
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    lol i know i got it confused with another problem. I was really good at algebra but this stuff is just killing me...

    • one year ago
  13. helpplease1 Group Title
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    hahaha no ph.d student here

    • one year ago
  14. looser66 Group Title
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    so? everything is quite simple, right?

    • one year ago
  15. helpplease1 Group Title
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    so i have \[2x ^{2}+3x-5=0\] and now factor? im sorry this class has confused me so much

    • one year ago
  16. looser66 Group Title
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    if you know how to factor, do it. if not, use discriminant. Either way is ok.

    • one year ago
  17. helpplease1 Group Title
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    x=-5/2 and x=1?

    • one year ago
  18. looser66 Group Title
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    yes, and reject -5/2, out of interval

    • one year ago
  19. helpplease1 Group Title
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    okay. so: cos(theta)=1? how do i make the general formula in form of of: _+_k

    • one year ago
  20. looser66 Group Title
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    not finish yet, cos \(\theta\) =1 \(\rightarrow \theta\) = 0 + 2k\(\pi\) that 's the general formula solution for you

    • one year ago
  21. helpplease1 Group Title
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    so the only solution is 0? because subbing 1 for k would make 2pi which would not be included in the interval?

    • one year ago
  22. looser66 Group Title
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    yes, you are right. general form is 2kpi, specific solution in the interval is 0 Medal for a good student

    • one year ago
  23. helpplease1 Group Title
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    oh my gosh thank you so much for your help!

    • one year ago
  24. looser66 Group Title
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    do I help? No! just you work on it, I did nothing. hehee

    • one year ago
  25. helpplease1 Group Title
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    :) :)

    • one year ago
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