## tanvidais13 Group Title 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

1. hartnn Group Title

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

2. tanvidais13 Group Title

isn't it??

3. hartnn Group Title

nopes

4. tanvidais13 Group Title

then what can i do here??

5. hartnn Group Title

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 Group Title

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

7. hartnn Group Title

am i making any sense?

8. hartnn Group Title

@estudier any ideas?

9. estudier Group Title

Just looking....

10. estudier Group Title

Where is the question from?

11. jasonxx Group Title

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

12. jasonxx Group Title

i mean cos^2 z

13. hartnn Group Title

trignometry only, i guess

14. jasonxx Group Title

sin^2 z=y

15. hartnn Group Title

how would that help @jasonxx ? what about sin 2x then?

16. jasonxx Group Title

it'll be a quadratic equation and it is done

17. jasonxx Group Title

something like 2y^2.....

18. hartnn Group Title

quadratic?? u have sin2x not sin^2 x

19. jasonxx Group Title

wait a sec guys ..we all are confused the question is

20. jasonxx Group Title

$2 \sin^4 z+ 7 \cos z=4$

21. jasonxx Group Title

$\cos z= \sqrt {1- \sin^2 z}$ let's say $\sin^2 z= y$ now i think it can be done

22. hartnn Group Title

u just increased the complexity by bringing square root sign....

23. jasonxx Group Title

$2y^2+7(\sqrt{1-y})=4$ $2y^2-4=7 \sqrt{1-y}$ square both sides and then let's say y^2=z

24. jasonxx Group Title

yea ..there could be alternative

25. hartnn Group Title

so u will get the same equation with y^4 which i hace posted

26. hartnn Group Title

2-4y^2+2y^4+7y=4

27. jasonxx Group Title

$4y^4-8y^2+16=49(1-y)$

28. hartnn Group Title

even that is not any simpler to solve

29. jasonxx Group Title

@tanvidais13 stuck or done lol, because you seem to be frozen ;)

30. tanvidais13 Group Title

sorry, guests over, had to "entertain" :P

31. Coolsector Group Title

are you sure you wrote the question correctly ?

32. jasonxx Group Title

no it is done wait a sec

33. jasonxx Group Title

$\sin^4 z= (\sin^2z)^2= (1-\cos^2z)^2$ now $1-2\cos^2z+\cos^4z=(1-\cos^2z)^2$

34. tanvidais13 Group Title

what jasonxx wrote in the beginning is right.

35. jasonxx Group Title

@tanvidais13 hope i entertained lol

36. hartnn Group Title

didn't entertain, didn't solve.... and the last thing u did,was the 1st thing i did

37. Coolsector Group Title

you are writing the same thing again

38. tanvidais13 Group Title

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 Group Title

@hartnn i know you might be fillipn out..relax mate, try to see things from everyone's visual angel

40. Coolsector Group Title

(a-b)^2 = a^2 -2ab + b^2

41. tanvidais13 Group Title

oh yeahh, so that was just the expansion.

42. hartnn Group Title

$$sin^2 x+cos^2 x =1$$and that expansion

43. jasonxx Group Title

@tanvidais13 i wrote sin^2 z = 1- cos^ z and that was an expansion

44. jasonxx Group Title

$(1- \cos^2 z)= \sin^2 z$ and $(1-\cos^2z)^2= expansion (a-b)^2$ ;)

45. tanvidais13 Group Title

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

46. jasonxx Group Title

yes you can...with a basic rearrangement

47. tanvidais13 Group Title

ENTERTAIN ME :P

48. jasonxx Group Title

are you sure ? lol

49. jasonxx Group Title

$\cos^4z-2\cos^2z+1+7cosz=4$ now can you play with this equation , see something is common among them, which can be shotned

50. jasonxx Group Title

wait a sec

51. jasonxx Group Title

$2 \cos^4z-4\cos^2z+2+7\cos z =4$

52. mathslover Group Title

@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 Group Title

Do refer : http://openstudy.com/code-of-conduct

54. jasonxx Group Title

@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 Group Title

if we say cos(z) = y, then we will have $2y^4 - 4y^2 + 7y - 2 = 0$

56. jasonxx Group Title

@tanvidais13 $2y^4-4y^2=2-7y$ now? can you do the rest

57. mathslover Group Title

@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 Group Title

see my 4th comment, 35 minutes ago, we are back there, we wasted 35 minutes..

59. jasonxx Group Title

@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 Group Title

Thanks and sorry if i was rude, but now someone help @tanvidais13 :)

61. jasonxx Group Title

@Mikael have a look, i believe you can take us outta this

62. jasonxx Group Title

@tanvidais13 if we take value of Z=0 degrees ... this equation fails !! are you sure this question is correct?

63. jasonxx Group Title

2*0+7*1 can't be equal to 4 ?? isn't it?

64. jasonxx Group Title

@mathslover what say bro? :) any idea !!

65. mathslover Group Title

not yet..

66. mathslover Group Title

@sauravshakya and @UnkleRhaukus

67. jasonxx Group Title

doesn't it look like equation that is not consistent ..it fails whilst substituting the value of z

68. Coolsector Group Title

but you had to find z .. you dont plug any value you want

69. jasonxx Group Title

good point, but how one is supposed to get value of somethin from (not sure) an incorrect equation

70. Coolsector Group Title

who said that it is incorrect

71. jasonxx Group Title

intuition ...let me solve this again :)

72. sauravshakya Group Title

Cant this be factored:|dw:1348921637348:dw|

73. jasonxx Group Title

no

74. siddhantsharan Group Title

We have to find roots?Or do we have to find number of roots?

75. sauravshakya Group Title

I think roots.

76. siddhantsharan Group Title

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 Group Title

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 Group Title

but unfortunately, we cannot use a calculator here....

79. sauravshakya Group Title

So, it has only two real solutions.

80. siddhantsharan Group Title

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

81. sauravshakya Group Title

|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 Group Title

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 Group Title

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.