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baseballguy1322 Group Title

Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360°

  • 11 months ago
  • 11 months ago

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  1. jdoe0001 Group Title
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    can you take common factor on the left-hand side?

    • 11 months ago
  2. baseballguy1322 Group Title
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    common factor would be sin(x) right?

    • 11 months ago
  3. jdoe0001 Group Title
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    ... h... actually... after getting rid of the 1, yes

    • 11 months ago
  4. jdoe0001 Group Title
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    well... 2sin(x)

    • 11 months ago
  5. baseballguy1322 Group Title
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    You add 1 to the right side but what would the left hand side look like after you simplify the gcf

    • 11 months ago
  6. jdoe0001 Group Title
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    \(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\ 4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\ 2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)

    • 11 months ago
  7. jdoe0001 Group Title
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    hmmmm

    • 11 months ago
  8. myininaya Group Title
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    So are you looking to solve 4u^3+2u^2-2u-1=0 Did you try to find the possible 0's?

    • 11 months ago
  9. jdoe0001 Group Title
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    yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side

    • 11 months ago
  10. myininaya Group Title
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    or you could factor by grouping :) much simpler

    • 11 months ago
  11. baseballguy1322 Group Title
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    Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?

    • 11 months ago
  12. myininaya Group Title
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    \[(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0\]

    • 11 months ago
  13. myininaya Group Title
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    Look at those first two terms. They have common factor. Look at the last terms. Factor out -1.

    • 11 months ago
  14. myininaya Group Title
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    Or you can look at it in terms of u instead of sin at first. Your choice.

    • 11 months ago
  15. baseballguy1322 Group Title
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    would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?

    • 11 months ago
  16. myininaya Group Title
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    \[(4u^3+2u^2)+(-2u-1)=0\]

    • 11 months ago
  17. myininaya Group Title
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    What does 4u^3 and 2u^2 have in common?

    • 11 months ago
  18. baseballguy1322 Group Title
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    2u^2? or just u^2?

    • 11 months ago
  19. myininaya Group Title
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    The gcf would be 2u^2 So we have \[2u^2(2u+1)+(-2u-1)=0\] Those last two terms, you can factor out -1.

    • 11 months ago
  20. baseballguy1322 Group Title
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    2u^2(2u+1) + ( 2sin) = 0

    • 11 months ago
  21. myininaya Group Title
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    (-2u-1)=-(2u+1)

    • 11 months ago
  22. myininaya Group Title
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    So (-2sin(x)-1)=-(2sin(x)+1)

    • 11 months ago
  23. myininaya Group Title
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    \[2u^2(2u+1)-1(2u+1)=0\] Now you have two terms. (2u+1) is a factor of both of those terms. Do you think you can completely factor the left hand side now.

    • 11 months ago
  24. baseballguy1322 Group Title
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    2u^2-1=0

    • 11 months ago
  25. myininaya Group Title
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    Don't forget about the (2u+1) So to factor \[2u^2(2u+1)-1(2u+1) \text{ we find what both of our terms have in common } \] The (2u+1) So we factor the (2u+1) out giving us \[(2u+1)(2u^2-1)=0\] Set both factors equal to 0. Don't forget u is sin(x).

    • 11 months ago
  26. baseballguy1322 Group Title
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    Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?

    • 11 months ago
  27. myininaya Group Title
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    Setting both factors equal to 0.

    • 11 months ago
  28. baseballguy1322 Group Title
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    sin x +1 = 0 sin x = -1 and sin^2x-1=0 sin^2x = 1

    • 11 months ago
  29. myininaya Group Title
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    Your 2's are disappearing?

    • 11 months ago
  30. myininaya Group Title
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    \[2u+1=0 \text{ or } 2u^2-1=0\] \[2u=-1 \text{ or } 2u^2=1\] \[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]

    • 11 months ago
  31. baseballguy1322 Group Title
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    How would you convert that to degrees for the final answer?

    • 11 months ago
  32. myininaya Group Title
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    sin(x)=-1/2 Did I just show you the unit circle?

    • 11 months ago
  33. baseballguy1322 Group Title
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    is it 210 and 330 degrees?

    • 11 months ago
  34. myininaya Group Title
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    You also need to solve sin^2(x)=1/2 Take the square root of both sides \[\sin(x)=\pm \sqrt{ \frac{1}{2} }\] Don't forget to solve these two as well. And yes those are the solutions to sin(x)=-1/2

    • 11 months ago
  35. myininaya Group Title
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    Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.

    • 11 months ago
  36. baseballguy1322 Group Title
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    so the four answers would be 30, 150, 210, and 330 degrees?

    • 11 months ago
  37. myininaya Group Title
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    \[\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}\] I don't think you solved these correctly.

    • 11 months ago
  38. myininaya Group Title
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    You solved sin(x)=-1/2 correctly. So part of your answer is 210 and 330 degrees.

    • 11 months ago
  39. baseballguy1322 Group Title
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    okay thanks for the help

    • 11 months ago
  40. myininaya Group Title
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    Np. You got it from here?

    • 11 months ago
  41. baseballguy1322 Group Title
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    yeah i got it.

    • 11 months ago
  42. myininaya Group Title
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    Neatness.

    • 11 months ago
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