Correct Answer - C
We have `sqrt(3) cos theta-3 sin theta=2(sin 5 theta - sin theta)`
or `(sqrt(3)//2) cos theta-(1//2) sin theta = sin 5 theta`
or `cos (theta+pi//6)= sin 5 theta=cos (pi//2-5 theta)`
`rArr theta+pi//6=2n pi pm (pi//2-5 theta), n in Z`
`rArr theta=(n pi//3)+(pi//18)`
or `theta=(-n pi//2)+(pi//6), AA n in Z`