## anonymous 3 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??

1. hartnn

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

2. anonymous

isn't it??

3. hartnn

nopes

4. anonymous

then what can i do here??

5. hartnn

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

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

7. hartnn

am i making any sense?

8. hartnn

@estudier any ideas?

9. anonymous

Just looking....

10. anonymous

Where is the question from?

11. anonymous

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

12. anonymous

i mean cos^2 z

13. hartnn

trignometry only, i guess

14. anonymous

sin^2 z=y

15. hartnn

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

16. anonymous

it'll be a quadratic equation and it is done

17. anonymous

something like 2y^2.....

18. hartnn

quadratic?? u have sin2x not sin^2 x

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. hartnn

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

23. anonymous

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

24. anonymous

yea ..there could be alternative

25. hartnn

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

26. hartnn

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

27. anonymous

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

28. hartnn

even that is not any simpler to solve

29. anonymous

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

30. anonymous

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

31. anonymous

are you sure you wrote the question correctly ?

32. anonymous

no it is done wait a sec

33. anonymous

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

34. anonymous

what jasonxx wrote in the beginning is right.

35. anonymous

@tanvidais13 hope i entertained lol

36. hartnn

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

37. anonymous

you are writing the same thing again

38. anonymous

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. anonymous

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

40. anonymous

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

41. anonymous

oh yeahh, so that was just the expansion.

42. hartnn

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

43. anonymous

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

44. anonymous

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

45. anonymous

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

46. anonymous

yes you can...with a basic rearrangement

47. anonymous

ENTERTAIN ME :P

48. anonymous

are you sure ? lol

49. anonymous

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

50. anonymous

wait a sec

51. anonymous

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

52. mathslover

@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

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

54. anonymous

@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. anonymous

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

56. anonymous

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

57. mathslover

58. hartnn

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

59. anonymous

@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

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

61. anonymous

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

62. anonymous

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

63. anonymous

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

64. anonymous

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

65. mathslover

not yet..

66. mathslover

@sauravshakya and @UnkleRhaukus

67. anonymous

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

68. anonymous

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

69. anonymous

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

70. anonymous

who said that it is incorrect

71. anonymous

intuition ...let me solve this again :)

72. anonymous

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

73. anonymous

no

74. anonymous

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

75. anonymous

I think roots.

76. anonymous

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

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

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

79. anonymous

So, it has only two real solutions.

80. anonymous

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

81. anonymous

|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. anonymous

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. anonymous

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.