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

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

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