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tanvidais13
Group Title
2sin^4(z) + 7cos(z) = 4, between 0 and 360 inclusive.
i could turn sin^4(z) into 1cos^4(z), and then turn that into (1+cos^2(z))(1cos^2(z)), but then what??
 one year ago
 one year ago
tanvidais13 Group Title
2sin^4(z) + 7cos(z) = 4, between 0 and 360 inclusive. i could turn sin^4(z) into 1cos^4(z), and then turn that into (1+cos^2(z))(1cos^2(z)), but then what??
 one year ago
 one year ago

This Question is Closed

hartnn Group TitleBest ResponseYou've already chosen the best response.3
how is sin^4(z) = 1cos^4(z) ??
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
isn't it??
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
then what can i do here??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
maybe (1cos^2 x)^2 = (1 2 cos^2 x + cos^4 x) (i have used sin^2 + cos^2= 1, so sin^2 = 1cos^2)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
24y^2+2y^4+7y=4.......by putting y=cos x not easy to solve, can u use calculator?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
am i making any sense?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
@estudier any ideas?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Just looking....
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Where is the question from?
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
cos z= (1+ sin 2z)/2 and let's say sin z= y ..you are done
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
i mean cos^2 z
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
trignometry only, i guess
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
how would that help @jasonxx ? what about sin 2x then?
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
it'll be a quadratic equation and it is done
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
something like 2y^2.....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
quadratic?? u have sin2x not sin^2 x
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
wait a sec guys ..we all are confused the question is
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[2 \sin^4 z+ 7 \cos z=4\]
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[\cos z= \sqrt {1 \sin^2 z}\] let's say \[\sin^2 z= y\] now i think it can be done
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
u just increased the complexity by bringing square root sign....
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[2y^2+7(\sqrt{1y})=4\] \[2y^24=7 \sqrt{1y}\] square both sides and then let's say y^2=z
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
yea ..there could be alternative
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
so u will get the same equation with y^4 which i hace posted
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
24y^2+2y^4+7y=4
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[4y^48y^2+16=49(1y)\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
even that is not any simpler to solve
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@tanvidais13 stuck or done lol, because you seem to be frozen ;)
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
sorry, guests over, had to "entertain" :P
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
are you sure you wrote the question correctly ?
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
no it is done wait a sec
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[\sin^4 z= (\sin^2z)^2= (1\cos^2z)^2\] now \[12\cos^2z+\cos^4z=(1\cos^2z)^2\]
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
what jasonxx wrote in the beginning is right.
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@tanvidais13 hope i entertained lol
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
didn't entertain, didn't solve.... and the last thing u did,was the 1st thing i did
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
you are writing the same thing again
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
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

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@hartnn i know you might be fillipn out..relax mate, try to see things from everyone's visual angel
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
(ab)^2 = a^2 2ab + b^2
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
oh yeahh, so that was just the expansion.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
\(sin^2 x+cos^2 x =1\)and that expansion
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@tanvidais13 i wrote sin^2 z = 1 cos^ z and that was an expansion
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[(1 \cos^2 z)= \sin^2 z\] and \[(1\cos^2z)^2= expansion (ab)^2\] ;)
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
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

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
yes you can...with a basic rearrangement
 one year ago

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
ENTERTAIN ME :P
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
are you sure ? lol
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[\cos^4z2\cos^2z+1+7cosz=4\] now can you play with this equation , see something is common among them, which can be shotned
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
wait a sec
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
\[2 \cos^4z4\cos^2z+2+7\cos z =4\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.2
@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

mathslover Group TitleBest ResponseYou've already chosen the best response.2
Do refer : http://openstudy.com/codeofconduct
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@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

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
if we say cos(z) = y, then we will have \[2y^4  4y^2 + 7y  2 = 0\]
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@tanvidais13 \[2y^44y^2=27y\] now? can you do the rest
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.2
@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 : 5x7=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

hartnn Group TitleBest ResponseYou've already chosen the best response.3
see my 4th comment, 35 minutes ago, we are back there, we wasted 35 minutes..
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@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

mathslover Group TitleBest ResponseYou've already chosen the best response.2
Thanks and sorry if i was rude, but now someone help @tanvidais13 :)
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@Mikael have a look, i believe you can take us outta this
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@tanvidais13 if we take value of Z=0 degrees ... this equation fails !! are you sure this question is correct?
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
2*0+7*1 can't be equal to 4 ?? isn't it?
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
@mathslover what say bro? :) any idea !!
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.2
not yet..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.2
@sauravshakya and @UnkleRhaukus
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
doesn't it look like equation that is not consistent ..it fails whilst substituting the value of z
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
but you had to find z .. you dont plug any value you want
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
good point, but how one is supposed to get value of somethin from (not sure) an incorrect equation
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
who said that it is incorrect
 one year ago

jasonxx Group TitleBest ResponseYou've already chosen the best response.1
intuition ...let me solve this again :)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Cant this be factored:dw:1348921637348:dw
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.0
We have to find roots?Or do we have to find number of roots?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
I think roots.
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.0
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

hartnn Group TitleBest ResponseYou've already chosen the best response.3
http://www.wolframalpha.com/input/?i=24y%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

hartnn Group TitleBest ResponseYou've already chosen the best response.3
but unfortunately, we cannot use a calculator here....
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
So, it has only two real solutions.
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.0
Precisely. And the other real root (not 2) is Probably not possible to compute by pen and paper.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
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

tanvidais13 Group TitleBest ResponseYou've already chosen the best response.2
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 strangeretricehellofaproblem :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

estudier Group TitleBest ResponseYou've already chosen the best response.0
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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