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baseballguy1322

  • one year ago

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

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  1. jdoe0001
    • one year ago
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    can you take common factor on the left-hand side?

  2. baseballguy1322
    • one year ago
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    common factor would be sin(x) right?

  3. jdoe0001
    • one year ago
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    ... h... actually... after getting rid of the 1, yes

  4. jdoe0001
    • one year ago
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    well... 2sin(x)

  5. baseballguy1322
    • one year ago
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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

  6. jdoe0001
    • one year ago
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    \(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\ 4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\ 2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)

  7. jdoe0001
    • one year ago
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    hmmmm

  8. myininaya
    • one year ago
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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?

  9. jdoe0001
    • one year ago
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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

  10. myininaya
    • one year ago
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    or you could factor by grouping :) much simpler

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

  12. myininaya
    • one year ago
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    \[(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0\]

  13. myininaya
    • one year ago
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    Look at those first two terms. They have common factor. Look at the last terms. Factor out -1.

  14. myininaya
    • one year ago
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    Or you can look at it in terms of u instead of sin at first. Your choice.

  15. baseballguy1322
    • one year ago
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    would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?

  16. myininaya
    • one year ago
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    \[(4u^3+2u^2)+(-2u-1)=0\]

  17. myininaya
    • one year ago
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    What does 4u^3 and 2u^2 have in common?

  18. baseballguy1322
    • one year ago
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    2u^2? or just u^2?

  19. myininaya
    • one year ago
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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.

  20. baseballguy1322
    • one year ago
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    2u^2(2u+1) + ( 2sin) = 0

  21. myininaya
    • one year ago
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    (-2u-1)=-(2u+1)

  22. myininaya
    • one year ago
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    So (-2sin(x)-1)=-(2sin(x)+1)

  23. myininaya
    • one year ago
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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.

  24. baseballguy1322
    • one year ago
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    2u^2-1=0

  25. myininaya
    • one year ago
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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).

  26. baseballguy1322
    • one year ago
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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?

  27. myininaya
    • one year ago
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    Setting both factors equal to 0.

  28. baseballguy1322
    • one year ago
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    sin x +1 = 0 sin x = -1 and sin^2x-1=0 sin^2x = 1

  29. myininaya
    • one year ago
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    Your 2's are disappearing?

  30. myininaya
    • one year ago
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    \[2u+1=0 \text{ or } 2u^2-1=0\] \[2u=-1 \text{ or } 2u^2=1\] \[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]

  31. baseballguy1322
    • one year ago
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    How would you convert that to degrees for the final answer?

  32. myininaya
    • one year ago
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    sin(x)=-1/2 Did I just show you the unit circle?

  33. baseballguy1322
    • one year ago
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    is it 210 and 330 degrees?

  34. myininaya
    • one year ago
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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

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

  36. baseballguy1322
    • one year ago
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    so the four answers would be 30, 150, 210, and 330 degrees?

  37. myininaya
    • one year ago
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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.

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

  39. baseballguy1322
    • one year ago
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    okay thanks for the help

  40. myininaya
    • one year ago
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    Np. You got it from here?

  41. baseballguy1322
    • one year ago
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    yeah i got it.

  42. myininaya
    • one year ago
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    Neatness.

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