1.

Let `S_(1)` be the set of all those solution of the equation `(1+a) cos theta cos(2 theta-b)=(1+a cos 2 theta) cos (theta -b)` which are independent of a and b and `S_(2)` be the set of all such solutions which are dependent on a and b. Then The set `S_(1)` and `S_(2)` areA. `{n pi, n in Z}` and `1/2 {n pi +(-1)^(n) sin^(-1) (a sin b) +b, n in Z}`B. `{n pi/2, n in Z}` and `{n pi+(-1)^(n) sin^(-1) (a sin b), n in Z}`C. `{n pi/2, n in Z}` and `{n pi +(-1)^(n) sin^(-1) (a/2 sin b), n in Z}`D. none of these

Answer» Correct Answer - A
`(1+a) cos theta cos (2 theta-b)=(1+a cos 2 theta) cos (theta-b)`
`rArr cos theta cos (2 theta-b)+a cos theta cos (2 theta-b)`
`= cos (theta-b) + a cos 2 theta cos (theta-b)`.
`rArr cos (3 theta -b)+ cos (theta-b)+a { cos (3 theta-b)+ cos (theta-b)}`
`=2 cos (theta-b)+a {cos (3 theta -b)+cos (theta+b)}`
`rArr cos (3 theta-b) - cos (theta-b)=a cos (theta+b)-a cos (theta-b)`
`rArr 2 sin (2 theta-b) sin theta =2 a sin theta sin b`
`rArr sin theta=0 or sin (2 theta-b) =a sin b`
if `sin theta=0` then `theta=n pi, n in Z rArr S_(1) = {n pi|n in Z|}`
Also, `2 theta-b=n pi +(-1)^(n) sin^(-1) (a sin b), n in Z`
`rArr S_(2)=1/2 {n pi +(-1)^(n) sin^(-1) (a sin b)+b, n in Z}`


Discussion

No Comment Found