anonymous
  • anonymous
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??
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

hartnn
  • hartnn
how is sin^4(z) = 1-cos^4(z) ??
anonymous
  • anonymous
isn't it??
hartnn
  • hartnn
nopes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
then what can i do here??
hartnn
  • 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)
hartnn
  • hartnn
2-4y^2+2y^4+7y=4.......by putting y=cos x not easy to solve, can u use calculator?
hartnn
  • hartnn
am i making any sense?
hartnn
  • hartnn
@estudier any ideas?
anonymous
  • anonymous
Just looking....
anonymous
  • anonymous
Where is the question from?
anonymous
  • anonymous
cos z= (1+ sin 2z)/2 and let's say sin z= y ..you are done
anonymous
  • anonymous
i mean cos^2 z
hartnn
  • hartnn
trignometry only, i guess
anonymous
  • anonymous
sin^2 z=y
hartnn
  • hartnn
how would that help @jasonxx ? what about sin 2x then?
anonymous
  • anonymous
it'll be a quadratic equation and it is done
anonymous
  • anonymous
something like 2y^2.....
hartnn
  • hartnn
quadratic?? u have sin2x not sin^2 x
anonymous
  • anonymous
wait a sec guys ..we all are confused the question is
anonymous
  • anonymous
\[2 \sin^4 z+ 7 \cos z=4\]
anonymous
  • anonymous
\[\cos z= \sqrt {1- \sin^2 z}\] let's say \[\sin^2 z= y\] now i think it can be done
hartnn
  • hartnn
u just increased the complexity by bringing square root sign....
anonymous
  • 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
anonymous
  • anonymous
yea ..there could be alternative
hartnn
  • hartnn
so u will get the same equation with y^4 which i hace posted
hartnn
  • hartnn
2-4y^2+2y^4+7y=4
anonymous
  • anonymous
\[4y^4-8y^2+16=49(1-y)\]
hartnn
  • hartnn
even that is not any simpler to solve
anonymous
  • anonymous
@tanvidais13 stuck or done lol, because you seem to be frozen ;)
anonymous
  • anonymous
sorry, guests over, had to "entertain" :P
anonymous
  • anonymous
are you sure you wrote the question correctly ?
anonymous
  • anonymous
no it is done wait a sec
anonymous
  • anonymous
\[\sin^4 z= (\sin^2z)^2= (1-\cos^2z)^2\] now \[1-2\cos^2z+\cos^4z=(1-\cos^2z)^2\]
anonymous
  • anonymous
what jasonxx wrote in the beginning is right.
anonymous
  • anonymous
@tanvidais13 hope i entertained lol
hartnn
  • hartnn
didn't entertain, didn't solve.... and the last thing u did,was the 1st thing i did
anonymous
  • anonymous
you are writing the same thing again
anonymous
  • 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???
anonymous
  • anonymous
@hartnn i know you might be fillipn out..relax mate, try to see things from everyone's visual angel
anonymous
  • anonymous
(a-b)^2 = a^2 -2ab + b^2
anonymous
  • anonymous
oh yeahh, so that was just the expansion.
hartnn
  • hartnn
\(sin^2 x+cos^2 x =1\)and that expansion
anonymous
  • anonymous
@tanvidais13 i wrote sin^2 z = 1- cos^ z and that was an expansion
anonymous
  • anonymous
\[(1- \cos^2 z)= \sin^2 z\] and \[(1-\cos^2z)^2= expansion (a-b)^2\] ;)
anonymous
  • anonymous
but then we only get cosz in powers of 1,2, and 4, not three. can we still solve it like that??
anonymous
  • anonymous
yes you can...with a basic rearrangement
anonymous
  • anonymous
ENTERTAIN ME :P
anonymous
  • anonymous
are you sure ? lol
anonymous
  • 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
anonymous
  • anonymous
wait a sec
anonymous
  • anonymous
\[2 \cos^4z-4\cos^2z+2+7\cos z =4\]
mathslover
  • 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.
mathslover
  • mathslover
anonymous
  • 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
anonymous
  • anonymous
if we say cos(z) = y, then we will have \[2y^4 - 4y^2 + 7y - 2 = 0\]
anonymous
  • anonymous
@tanvidais13 \[2y^4-4y^2=2-7y\] now? can you do the rest
mathslover
  • mathslover
@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 :)
hartnn
  • hartnn
see my 4th comment, 35 minutes ago, we are back there, we wasted 35 minutes..
anonymous
  • 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
mathslover
  • mathslover
Thanks and sorry if i was rude, but now someone help @tanvidais13 :)
anonymous
  • anonymous
@Mikael have a look, i believe you can take us outta this
anonymous
  • anonymous
@tanvidais13 if we take value of Z=0 degrees ... this equation fails !! are you sure this question is correct?
anonymous
  • anonymous
2*0+7*1 can't be equal to 4 ?? isn't it?
anonymous
  • anonymous
@mathslover what say bro? :) any idea !!
mathslover
  • mathslover
not yet..
mathslover
  • mathslover
anonymous
  • anonymous
doesn't it look like equation that is not consistent ..it fails whilst substituting the value of z
anonymous
  • anonymous
but you had to find z .. you dont plug any value you want
anonymous
  • anonymous
good point, but how one is supposed to get value of somethin from (not sure) an incorrect equation
anonymous
  • anonymous
who said that it is incorrect
anonymous
  • anonymous
intuition ...let me solve this again :)
anonymous
  • anonymous
Cant this be factored:|dw:1348921637348:dw|
anonymous
  • anonymous
no
anonymous
  • anonymous
We have to find roots?Or do we have to find number of roots?
anonymous
  • anonymous
I think roots.
anonymous
  • 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
hartnn
  • 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
hartnn
  • hartnn
but unfortunately, we cannot use a calculator here....
anonymous
  • anonymous
So, it has only two real solutions.
anonymous
  • anonymous
Precisely. And the other real root (not -2) is Probably not possible to compute by pen and paper.
anonymous
  • 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.
anonymous
  • 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
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.