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tanvidais13

  • 2 years ago

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??

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  1. hartnn
    • 2 years ago
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    how is sin^4(z) = 1-cos^4(z) ??

  2. tanvidais13
    • 2 years ago
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    isn't it??

  3. hartnn
    • 2 years ago
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    nopes

  4. tanvidais13
    • 2 years ago
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    then what can i do here??

  5. hartnn
    • 2 years ago
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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)

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

  7. hartnn
    • 2 years ago
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    am i making any sense?

  8. hartnn
    • 2 years ago
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    @estudier any ideas?

  9. estudier
    • 2 years ago
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    Just looking....

  10. estudier
    • 2 years ago
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    Where is the question from?

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

  12. jasonxx
    • 2 years ago
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    i mean cos^2 z

  13. hartnn
    • 2 years ago
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    trignometry only, i guess

  14. jasonxx
    • 2 years ago
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    sin^2 z=y

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

  16. jasonxx
    • 2 years ago
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    it'll be a quadratic equation and it is done

  17. jasonxx
    • 2 years ago
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    something like 2y^2.....

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

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

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

  21. jasonxx
    • 2 years ago
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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

  22. hartnn
    • 2 years ago
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    u just increased the complexity by bringing square root sign....

  23. jasonxx
    • 2 years ago
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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

  24. jasonxx
    • 2 years ago
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    yea ..there could be alternative

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

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

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

  28. hartnn
    • 2 years ago
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    even that is not any simpler to solve

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

  30. tanvidais13
    • 2 years ago
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    sorry, guests over, had to "entertain" :P

  31. Coolsector
    • 2 years ago
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    are you sure you wrote the question correctly ?

  32. jasonxx
    • 2 years ago
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    no it is done wait a sec

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

  34. tanvidais13
    • 2 years ago
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    what jasonxx wrote in the beginning is right.

  35. jasonxx
    • 2 years ago
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    @tanvidais13 hope i entertained lol

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

  37. Coolsector
    • 2 years ago
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    you are writing the same thing again

  38. tanvidais13
    • 2 years ago
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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???

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

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

  41. tanvidais13
    • 2 years ago
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    oh yeahh, so that was just the expansion.

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

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

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

  45. tanvidais13
    • 2 years ago
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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??

  46. jasonxx
    • 2 years ago
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    yes you can...with a basic rearrangement

  47. tanvidais13
    • 2 years ago
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    ENTERTAIN ME :P

  48. jasonxx
    • 2 years ago
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    are you sure ? lol

  49. jasonxx
    • 2 years ago
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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

  50. jasonxx
    • 2 years ago
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    wait a sec

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

  52. mathslover
    • 2 years ago
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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.

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

  54. jasonxx
    • 2 years ago
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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

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

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

  57. mathslover
    • 2 years ago
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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 :)

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

  59. jasonxx
    • 2 years ago
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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

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

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

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

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

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

  65. mathslover
    • 2 years ago
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    not yet..

  66. mathslover
    • 2 years ago
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    @sauravshakya and @UnkleRhaukus

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

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

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

  70. Coolsector
    • 2 years ago
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    who said that it is incorrect

  71. jasonxx
    • 2 years ago
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    intuition ...let me solve this again :)

  72. sauravshakya
    • 2 years ago
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    Cant this be factored:|dw:1348921637348:dw|

  73. jasonxx
    • 2 years ago
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    no

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

  75. sauravshakya
    • 2 years ago
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    I think roots.

  76. siddhantsharan
    • 2 years ago
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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

  77. hartnn
    • 2 years ago
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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

  78. hartnn
    • 2 years ago
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    but unfortunately, we cannot use a calculator here....

  79. sauravshakya
    • 2 years ago
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    So, it has only two real solutions.

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

  81. sauravshakya
    • 2 years ago
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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.

  82. tanvidais13
    • 2 years ago
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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

  83. estudier
    • 2 years ago
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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.

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