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Lulu212 Group Title

Solve sin2x + 4sin x + 3 = 0

  • one year ago
  • one year ago

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  1. Lulu212 Group Title
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    do u know how to solve this ?

    • one year ago
  2. jeanlouie Group Title
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    sin(2x) is more interesting :)

    • one year ago
  3. Omniscience Group Title
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    Is it sin^2x + 4sinx + 3 = 0 ?

    • one year ago
  4. Omniscience Group Title
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    |dw:1345778720492:dw| i dont know why latex is not working..

    • one year ago
  5. ash2326 Group Title
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    @Lulu212 Is it \(\sin 2x \ or \ \sin^2 x\)?

    • one year ago
  6. Lulu212 Group Title
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    sin 2x

    • one year ago
  7. Mimi_x3 Group Title
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    \[sin2x + 4sinx + 3\]

    • one year ago
  8. Mimi_x3 Group Title
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    \[sin2x = 2sinxcosx\] \[ 2sinxcosx + 4sinx + 3 = 0 \]

    • one year ago
  9. Mimi_x3 Group Title
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    have you tried that yet?

    • one year ago
  10. Yahoo! Group Title
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    hm...hm.... u need...to find the General soln of this....!

    • one year ago
  11. Yahoo! Group Title
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    @UnkleRhaukus @Hero @jim_thompson5910 @myininaya

    • one year ago
  12. Yahoo! Group Title
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    @Hero @hartnn

    • one year ago
  13. UnkleRhaukus Group Title
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    \[ \sin(2x) + 4\sin (x) + 3 = 0\] \[ 2\sin(x)\cos(x) + 4\sin (x) + 3 = 0\] \[2\sin (x)\left(\cos(x)+2\right)+3=0\] \[2\sin (x)\left(\cos(x)+2\right)=-3\] \[\cos(x)+2=\frac{-3}2\csc(x)\] i dont know

    • one year ago
  14. Yahoo! Group Title
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    @satellite73 @saadi

    • one year ago
  15. Hero Group Title
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    I don't understand why @UnkleRhaukus can't finish it

    • one year ago
  16. Yahoo! Group Title
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    Can u Finish @Hero i am also stuck..there

    • one year ago
  17. Yahoo! Group Title
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    @jim_thompson5910

    • one year ago
  18. Mimi_x3 Group Title
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    hmm http://www.wolframalpha.com/input/?i=sin2x+%2B+4sin+x+%2B+3+%3D+0 i dont think its possible..

    • one year ago
  19. Yahoo! Group Title
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    i think..tthere is mistake in question..))))

    • one year ago
  20. Yahoo! Group Title
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    @Lulu212 can u check ur question...))

    • one year ago
  21. Mimi_x3 Group Title
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    I think it should be \(sin^2x + 4sinx +3 =0\)

    • one year ago
  22. Mimi_x3 Group Title
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    then \(u=sinx\) more reasonable

    • one year ago
  23. Yahoo! Group Title
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    then it is a quad eq

    • one year ago
  24. Yahoo! Group Title
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    we can find sinx

    • one year ago
  25. Hero Group Title
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    I'm inclined to agree with @Mimi_x3

    • one year ago
  26. manishsatywali Group Title
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    this question can be solved ............why u guys are worrying????????

    • one year ago
  27. nickhouraney Group Title
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    it cant be solved just endless ways of writing the equation

    • one year ago
  28. panlac01 Group Title
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    what are we solving for? LOL. that equation needs to be written precisely in a way that we all can agree to what it is.

    • one year ago
  29. UnkleRhaukus Group Title
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    \[\sin^2x+4\sin x+3=0\] \[u^2-4u+3=0\] \[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\] \[u=2\pm\frac{\sqrt{4}}{2}\]\[u=2\pm1\]\[u=1,3\] \[(1)^2+4(1)+3=0,\qquad\qquad 3^2-4(3)+3=0\]\[8=0,\qquad\qquad\qquad 3^2-4(3)+3=0\]\[\text{false}\qquad\qquad\qquad 9-12+3=0\]\[\qquad\qquad\qquad\qquad0=0\]\[\qquad\qquad\qquad\text{true}\]

    • one year ago
  30. panlac01 Group Title
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    the equation toolbar in this thing sucks hairy ______. how do you write it so neatly like that?

    • one year ago
  31. UnkleRhaukus Group Title
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    so \(u=\sin(x)=3\) \[x=\arcsin (3)\] ... this question is very confusing \[\text{\[u=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(3)}}{2(1)}\]}\]

    • one year ago
  32. UnkleRhaukus Group Title
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    oh damn my solution is way off \[\sin^2(\arcsin 3) + 4\sin (\arcsin 3) + 3 = 24≠0\]

    • one year ago
  33. UnkleRhaukus Group Title
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    oh \(u^2+4u+3=0\) NOT \(u^2-4u+3=0\)

    • one year ago
  34. nickhouraney Group Title
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    but this wasnt the problem that was stated lol

    • one year ago
  35. UnkleRhaukus Group Title
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    \[u_{1,2}=-2\pm1=-3,-1\]

    • one year ago
  36. UnkleRhaukus Group Title
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    \[\sin x=u=-3,-1\] \[x_{1,2}=\arcsin (-1),\quad \arcsin(-3)\] \[x_1=-\frac \pi2\qquad, x_2=\text{some transcendental nonsense} \] hence\[x=-\frac {\pi}2\] CHECKING \[\sin^2x + 4\sin x + 3 = 0\] \[\sin^2\left(-\frac\pi 2\right) + 4\sin \left(-\frac\pi 2\right) + 3 = 0\]\[1+4(-1)+3=0\]\[0=0\]true hence \[\large{x=-\frac {\pi}2}\]is the solution to \[\sin^2x + 4\sin x + 3 = 0\]

    • one year ago
  37. UnkleRhaukus Group Title
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    ta-da

    • one year ago
  38. Hero Group Title
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    People who post questions need to learn how to use the equation editors to avoid such confusion.

    • one year ago
  39. panlac01 Group Title
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    couldn't agree more @Hero

    • one year ago
  40. UnkleRhaukus Group Title
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    i think someone needs to write a program that can interpet questions and convert them to \(\LaTeX\) automagically

    • one year ago
  41. nickhouraney Group Title
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    @lulu212 i am disappoint at this equation

    • one year ago
  42. Hero Group Title
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    I imagine that something like that would have to be interactive. It would have to be some kind of AI that would automatically ask the user something like: Did you mean \(\sin(2x)\) or \(\sin^2(x)\)?

    • one year ago
  43. panlac01 Group Title
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    beyond OS' capacity :P we can ask Wolfram though. That dude is probably the smartest programmer I've seen alive

    • one year ago
  44. Hero Group Title
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    That's because he was reincarnated Einstein

    • one year ago
  45. panlac01 Group Title
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    hahaha... I think he was already born when Einstein was still alive

    • one year ago
  46. Hero Group Title
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    Well, Einstein's lost twin

    • one year ago
  47. nickhouraney Group Title
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    this question has like 40 responses all because of a typing error

    • one year ago
  48. Hero Group Title
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    We're still assuming that it is a typing error

    • one year ago
  49. nickhouraney Group Title
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    it is

    • one year ago
  50. Hero Group Title
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    There's no such thing as typing errors. Only human errors.

    • one year ago
  51. panlac01 Group Title
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    the person who asked for help did not even say anything after the question was posted (TYPOGRAPHICAL ERROR)

    • one year ago
  52. UnkleRhaukus Group Title
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    can't we all be einsteins lost twin?

    • one year ago
  53. panlac01 Group Title
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    I know that I can't... I have a roach's brain compare to his

    • one year ago
  54. nickhouraney Group Title
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    same difference hero lol

    • one year ago
  55. UnkleRhaukus Group Title
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    twins dont necessarily have the same charistics

    • one year ago
  56. panlac01 Group Title
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    oh that's right... I'd be the opposite of his intellect then

    • one year ago
  57. UnkleRhaukus Group Title
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    * characteristic

    • one year ago
  58. Hero Group Title
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    In one of Einstein's lost journals, he wrote that his brain was removed as a kid. The person handling his brain accidentally dropped in brain in a toilet, but quietly removed it without telling anyone. Once the brain was placed back into Einstein's head, he felt different, like as if he had more 'powers' and 'ability'. We all know what that led to.

    • one year ago
  59. UnkleRhaukus Group Title
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    oh my.

    • one year ago
  60. panlac01 Group Title
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    where's David Hume?

    • one year ago
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