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

isn't it??

nopes

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?

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

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 1âˆ’2cos^2(z)+cos^4(z)=(1âˆ’cos^2(z))2
what is that???

(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

wait a sec

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

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

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

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

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

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.

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.