Question

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??

Answer 1

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

Answer 2

isn't it??

Answer 3

nopes

Answer 4

then what can i do here??

Answer 5

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)

Answer 6

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

Answer 7

am i making any sense?

Answer 8

@estudier any ideas?

Answer 9

Just looking....

Answer 10

Where is the question from?

Answer 11

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

Answer 12

i mean cos^2 z

Answer 13

trignometry only, i guess

Answer 14

sin^2 z=y

Answer 15

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

Answer 16

it'll be a quadratic equation and it is done

Answer 17

something like 2y^2.....

Answer 18

quadratic?? u have sin2x not sin^2 x

Answer 19

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

Answer 20

\[2 \sin^4 z+ 7 \cos z=4\]

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yea ..there could be alternative

Answer 25

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

Answer 26

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

Answer 27

\[4y^4-8y^2+16=49(1-y)\]

Answer 28

even that is not any simpler to solve

Answer 29

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

Answer 30

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

Answer 31

are you sure you wrote the question correctly ?

Answer 32

no it is done wait a sec

Answer 33

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

Answer 34

what jasonxx wrote in the beginning is right.

Answer 35

@tanvidais13 hope i entertained lol

Answer 36

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

Answer 37

you are writing the same thing again

Answer 38

wait i STILL didn't get how you got 1−2cos^2(z)+cos^4(z)=(1−cos^2(z))2

what is that???

Answer 39

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

Answer 40

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

Answer 41

oh yeahh, so that was just the expansion.

Answer 42

\(sin^2 x+cos^2 x =1\)and that expansion

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

yes you can...with a basic rearrangement

Answer 47

ENTERTAIN ME :P

Answer 48

are you sure ? lol

Answer 49

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

Answer 50

wait a sec

Answer 51

\[2 \cos^4z-4\cos^2z+2+7\cos z =4\]

Answer 52

@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.

Answer 53

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

Answer 54

@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

Answer 55

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

Answer 56

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

Answer 57

@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 :)

Answer 58

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

Answer 59

@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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

not yet..

Answer 66

@sauravshakya and @UnkleRhaukus

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

who said that it is incorrect

Answer 71

intuition ...let me solve this again :)

Answer 72

Cant this be factored: