But eureka when i actually solved it i got a i = âˆš1 in between how do i solve that
d arithmetic mean of d roots of d equatio n
4Cos^{3}x 4Cos^{2}x Cos[315pie+x]=1

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2nÏ€=x is the answer as range of x is [0,315] so convert 315 into radian 315Ã—Ï€/Ï€â‰ˆ100Ï€ so x=0,2Ï€,4Ï€......100Ï€ and atithimetic mean comes out to be (100Ï€+0)/2=50Ï€
the question is indeed wrong...........
right question is
the arithmetic mean of d roots of the equation
4Cos^{3}x 4Cos^{2}x Cos(Ï€+x)=1 in interval (0,315)
solution:
proceeding as nishant sir did,we get cosx=1
=>cosx=2n.Ï€
Since 100Ï€<315<101Ï€
=>cosx=2Ï€,4Ï€,...........100Ï€
=>AM=2(Ï€+2Ï€+........50Ï€)/50
=>AM=2*50*51*Ï€/2*50
=>AM=51Ï€
do you mean that you got the answer 51 pi?
I think that is the answer
can you tell the other options
and also can you tell the source of this question?
idont understand wat ur saying bhaiyya d ans is given as 51âˆ© i got it can ne 1 say how it is plz
4Cos3x 4Cos2x Cos[315pie+x]=1
=
4Cos^{3}x 4Cos^{2}x + Cosx1=0
( 4 cos^{2}x +1 ) (cos x  1) =0
the first one is not possible...
so x=(2n)Ï€
among the options given to your brinda... you should realise that no other option will hold.. I think you must be solving a multiple choice question..
u wrote .....
a = i/2
=> cosx=i/2
writing this doesnt mean anything
i hope u unerstand.......
it reduces to 4 cos^{3}x 4cos^{2}x+cosx1=o
take cos x1 common......
when u had till here......4Cos^{3}x  4 Cos^{2}x + Cosx = 1
then use the property.....
summation of roots=(4/4)=1
=> AM=1/3
4Cos^{3}x  4 Cos^{2}x + Cosx = 1
let Cos x = a
4a^{3}  4a^{2} + a = 1
4a^{3}  4a^{2} + a  1 = 0
(a  1)(4a^{2} + 1) = 0
Therefore possible
a = 1
a = i/2
A.M = 1 + i /3
AM = 1/3 + i/3