If the expression `([s in(x/2)+cos(x/2)-i t a n(x)])/([1+2is in(x/2)])` is real, then the set of all possible values of `x` is.........

Answer

x = `2 n pi + 2 alpha , alpha = tan^(-1) k ` , where k `in (1 ,2 )` or x = 2 `n pi`

Explanation

`((sin x/2+cos x/2)- I tan x )/(1+2 i sin x/2)in R`
`=((sin"" x/2+ cos""x/2-i tanx)(1-2 i sin ""x/2))/(1+4 sin^2""x/2)`
Since , it is real so imaginary part will be zero
`therefore -2 sin ""x/2(sin ""x/2+ cos ""x/2)-tan x=0`
`rArr -2 sin ""x/2(sin ""x/2+ cos ""x/2)cos x+2 sin ""x/2cos ""x/2=0`
`rArr sin ""x/2[(sin""x/2cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2]=0`
`therefore sin ""x/2=0`
`rArr x= 2 npi`
or `(sin ""x/2+cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2=0`
on dividing by `cos^3""x/2` we get
`(tan ""x/2+1)(1-tan^2""x/2)+(1+tan^2""x/2)=0`
`rArr tan^3""x/2-tan""x/2-2=0`
Let `tan""x/2=t`
and `f(t)=t^3-t-2`
then `f(1)=-2 lt 0`
and `f(2)=4 gt 0 `
Thus f(t) changes sign from negative to positve in the interval (1,2)
`therefore` Let t= k be the root for which
f(k)=0 and `k in (1,2)`
`therefore` t=k or `tan ""x/2=k = tan alpha`
`rArr x//2 = npi+alpha`
`rArr {:{(x=2npi+2alpha ","alpha=tan^(-1)k",where k" in "(1,2)"),(" "or x=2npi):}`