d arithmetic mean of d roots of d equatio n
4Cos3x -4Cos2x -Cos[315pie+x]=1
11But eureka when i actually solved it i got a i = √-1 in between how do i solve that
12nπ=x is the answer as range of x is [0,315] so convert 315 into radian 315×π/π≈100π so x=0,2π,4π......100π and atithimetic mean comes out to be (100π+0)/2=50π
24the question is indeed wrong...........
right question is
the arithmetic mean of d roots of the equation
4Cos3x -4Cos2x -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€
62do 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?
1idont understand wat ur saying bhaiyya d ans is given as 51∩ i got it can ne 1 say how it is plz
624Cos3x -4Cos2x -Cos[315pie+x]=1
=
4Cos3x -4Cos2x + Cosx-1=0
( 4 cos2x +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..
24u wrote .....
a = i/2
=> cosx=i/2
writing this doesnt mean anything
i hope u unerstand.......
1it reduces to 4 cos3x -4cos2x+cosx-1=o
take cos x-1 common......
24when u had till here......4Cos3x - 4 Cos2x + Cosx = 1
then use the property.....
summation of roots=-(-4/4)=1
=> AM=1/3
114Cos3x - 4 Cos2x + Cosx = 1
let Cos x = a
4a3 - 4a2 + a = 1
4a3 - 4a2 + a - 1 = 0
(a - 1)(4a2 + 1) = 0
Therefore possible
a = 1
a = i/2
A.M = 1 + i /3
AM = 1/3 + i/3
