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tanvidais13

2sin^4(z) + 7cos(z) = 4, between 0 and 360 inclusive. i could turn sin^4(z) into 1-cos^4(z), and then turn that into (1+cos^2(z))(1-cos^2(z)), but then what??

  • one year ago
  • one year ago

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  1. hartnn
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    how is sin^4(z) = 1-cos^4(z) ??

    • one year ago
  2. tanvidais13
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    isn't it??

    • one year ago
  3. hartnn
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    nopes

    • one year ago
  4. tanvidais13
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    then what can i do here??

    • one year ago
  5. hartnn
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    maybe (1-cos^2 x)^2 = (1- 2 cos^2 x + cos^4 x) (i have used sin^2 + cos^2= 1, so sin^2 = 1-cos^2)

    • one year ago
  6. hartnn
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    2-4y^2+2y^4+7y=4.......by putting y=cos x not easy to solve, can u use calculator?

    • one year ago
  7. hartnn
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    am i making any sense?

    • one year ago
  8. hartnn
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    @estudier any ideas?

    • one year ago
  9. estudier
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    Just looking....

    • one year ago
  10. estudier
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    Where is the question from?

    • one year ago
  11. jasonxx
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    cos z= (1+ sin 2z)/2 and let's say sin z= y ..you are done

    • one year ago
  12. jasonxx
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    i mean cos^2 z

    • one year ago
  13. hartnn
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    trignometry only, i guess

    • one year ago
  14. jasonxx
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    sin^2 z=y

    • one year ago
  15. hartnn
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    how would that help @jasonxx ? what about sin 2x then?

    • one year ago
  16. jasonxx
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    it'll be a quadratic equation and it is done

    • one year ago
  17. jasonxx
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    something like 2y^2.....

    • one year ago
  18. hartnn
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    quadratic?? u have sin2x not sin^2 x

    • one year ago
  19. jasonxx
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    wait a sec guys ..we all are confused the question is

    • one year ago
  20. jasonxx
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    \[2 \sin^4 z+ 7 \cos z=4\]

    • one year ago
  21. jasonxx
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    \[\cos z= \sqrt {1- \sin^2 z}\] let's say \[\sin^2 z= y\] now i think it can be done

    • one year ago
  22. hartnn
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    u just increased the complexity by bringing square root sign....

    • one year ago
  23. jasonxx
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    \[2y^2+7(\sqrt{1-y})=4\] \[2y^2-4=7 \sqrt{1-y}\] square both sides and then let's say y^2=z

    • one year ago
  24. jasonxx
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    yea ..there could be alternative

    • one year ago
  25. hartnn
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    so u will get the same equation with y^4 which i hace posted

    • one year ago
  26. hartnn
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    2-4y^2+2y^4+7y=4

    • one year ago
  27. jasonxx
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    \[4y^4-8y^2+16=49(1-y)\]

    • one year ago
  28. hartnn
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    even that is not any simpler to solve

    • one year ago
  29. jasonxx
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    @tanvidais13 stuck or done lol, because you seem to be frozen ;)

    • one year ago
  30. tanvidais13
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    sorry, guests over, had to "entertain" :P

    • one year ago
  31. Coolsector
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    are you sure you wrote the question correctly ?

    • one year ago
  32. jasonxx
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    no it is done wait a sec

    • one year ago
  33. jasonxx
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    \[\sin^4 z= (\sin^2z)^2= (1-\cos^2z)^2\] now \[1-2\cos^2z+\cos^4z=(1-\cos^2z)^2\]

    • one year ago
  34. tanvidais13
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    what jasonxx wrote in the beginning is right.

    • one year ago
  35. jasonxx
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    @tanvidais13 hope i entertained lol

    • one year ago
  36. hartnn
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    didn't entertain, didn't solve.... and the last thing u did,was the 1st thing i did

    • one year ago
  37. Coolsector
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    you are writing the same thing again

    • one year ago
  38. tanvidais13
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    wait i STILL didn't get how you got 1−2cos^2(z)+cos^4(z)=(1−cos^2(z))2 what is that???

    • one year ago
  39. jasonxx
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    @hartnn i know you might be fillipn out..relax mate, try to see things from everyone's visual angel

    • one year ago
  40. Coolsector
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    (a-b)^2 = a^2 -2ab + b^2

    • one year ago
  41. tanvidais13
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    oh yeahh, so that was just the expansion.

    • one year ago
  42. hartnn
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    \(sin^2 x+cos^2 x =1\)and that expansion

    • one year ago
  43. jasonxx
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    @tanvidais13 i wrote sin^2 z = 1- cos^ z and that was an expansion

    • one year ago
  44. jasonxx
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    \[(1- \cos^2 z)= \sin^2 z\] and \[(1-\cos^2z)^2= expansion (a-b)^2\] ;)

    • one year ago
  45. tanvidais13
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    but then we only get cosz in powers of 1,2, and 4, not three. can we still solve it like that??

    • one year ago
  46. jasonxx
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    yes you can...with a basic rearrangement

    • one year ago
  47. tanvidais13
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    ENTERTAIN ME :P

    • one year ago
  48. jasonxx
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    are you sure ? lol

    • one year ago
  49. jasonxx
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    \[\cos^4z-2\cos^2z+1+7cosz=4\] now can you play with this equation , see something is common among them, which can be shotned

    • one year ago
  50. jasonxx
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    wait a sec

    • one year ago
  51. jasonxx
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    \[2 \cos^4z-4\cos^2z+2+7\cos z =4\]

    • one year ago
  52. mathslover
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    @jasonxx I request you to please don't interrupt when some one is helping a user. No worries, if you were ONLY giving suggestion but if you are spoiling the work of the previous user then it will not be nice.

    • one year ago
  53. mathslover
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    Do refer : http://openstudy.com/code-of-conduct

    • one year ago
  54. jasonxx
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    @mathslover i do respect your words but i can't stop myself from some arrogant person and his haughty comment i am sure you know, whom am i talking about

    • one year ago
  55. tanvidais13
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    if we say cos(z) = y, then we will have \[2y^4 - 4y^2 + 7y - 2 = 0\]

    • one year ago
  56. jasonxx
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    @tanvidais13 \[2y^4-4y^2=2-7y\] now? can you do the rest

    • one year ago
  57. mathslover
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    @jasonxx please stay friendly dude. It would be better if you have not used "arrogant" like words, this makes the whole difference. It's nice to see that you respect my words but I will say that even I am spoiling this post now, so I am leaving , you may continue your help but remember the following things from now onwards: i) Never give direct answer to any asker. ii) Don't give direct solution to the asker, this not only ends the motive of openstudy but also not helps asker. iii) Let the asker do the work, you must suggest him/her regarding your idea or steps. like : 5x-7=3 , just say "Try to add 7 both sides, what do you get?" iv) "Answer snipping" , this is what you have likely done here, if a user is helping the asker then please don't interrupt by giving direct solution or any way, but if you have good ideas then just post shortly. Thanks :)

    • one year ago
  58. hartnn
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    see my 4th comment, 35 minutes ago, we are back there, we wasted 35 minutes..

    • one year ago
  59. jasonxx
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    @mathslover thank you but you should not be biased..well for the sake of us i am putting this to an end, i haven't copied i was friendly and i was working the way i believe is dood and is in accordance with this code of conduct but if one is willing to have my welcome in some other way then i am sorry, either i'll help or i shall feel compelled to assault someone verbally ...my serious apology for being rude thank you

    • one year ago
  60. mathslover
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    Thanks and sorry if i was rude, but now someone help @tanvidais13 :)

    • one year ago
  61. jasonxx
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    @Mikael have a look, i believe you can take us outta this

    • one year ago
  62. jasonxx
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    @tanvidais13 if we take value of Z=0 degrees ... this equation fails !! are you sure this question is correct?

    • one year ago
  63. jasonxx
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    2*0+7*1 can't be equal to 4 ?? isn't it?

    • one year ago
  64. jasonxx
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    @mathslover what say bro? :) any idea !!

    • one year ago
  65. mathslover
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    not yet..

    • one year ago
  66. mathslover
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    @sauravshakya and @UnkleRhaukus

    • one year ago
  67. jasonxx
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    doesn't it look like equation that is not consistent ..it fails whilst substituting the value of z

    • one year ago
  68. Coolsector
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    but you had to find z .. you dont plug any value you want

    • one year ago
  69. jasonxx
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    good point, but how one is supposed to get value of somethin from (not sure) an incorrect equation

    • one year ago
  70. Coolsector
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    who said that it is incorrect

    • one year ago
  71. jasonxx
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    intuition ...let me solve this again :)

    • one year ago
  72. sauravshakya
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    Cant this be factored:|dw:1348921637348:dw|

    • one year ago
  73. jasonxx
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    no

    • one year ago
  74. siddhantsharan
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    We have to find roots?Or do we have to find number of roots?

    • one year ago
  75. sauravshakya
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    I think roots.

    • one year ago
  76. siddhantsharan
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    The question doesnt specify. I dont think you can find the exact root to this. http://www.wolframalpha.com/input/?i=2x%5E4+-+4x%5E2+%2B+7x+-+2

    • one year ago
  77. hartnn
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    http://www.wolframalpha.com/input/?i=2-4y%5E2%2B2y%5E4%2B7y%3D4 y=0.35 can be cos z, other two are imaginary and last is -2 cos z cannot be -2 so cos z=0.35

    • one year ago
  78. hartnn
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    but unfortunately, we cannot use a calculator here....

    • one year ago
  79. sauravshakya
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    So, it has only two real solutions.

    • one year ago
  80. siddhantsharan
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    Precisely. And the other real root (not -2) is Probably not possible to compute by pen and paper.

    • one year ago
  81. sauravshakya
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    |dw:1348922391902:dw| AND I was trying to prove it has no real solution as the above equation is never 0 for -1<=t<=1..... But I was doing all wrong.

    • one year ago
  82. tanvidais13
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    so the only possible answer is cos(z) = 0.35, only the problem is it's basically impossible to find out by hand :P anyways, THANK YOU SO MUCH, everyone, for this strange-retricehell-of-a-problem :P AND this includes all the relative latecomers :P I can't give a medal to everyone, but I wish I could :P

    • one year ago
  83. estudier
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    If you absolutely have to find it by hand and you can make an initial guess which is close enough, then you could perhaps use Newton's method to get the root.

    • one year ago
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