f(x)=cos−1(2[|sinx|+|cosx|]sin2x+2sinx+114)   ( [ ] denotes greatest integer function}. Then domain of f(x) in the internal [0, 2π] is.

Given that f(x)=cos−1(2[|sinx|+|cosx|]sin2x+2sinx+114) We have1≤|sinx|+|cosx|≤2[|sinx|+|cosx|]=1    ∀x∈R  Where [.]  is G.I.FNowsin2x+2sinx+114=(sinx+1)2+74For f to be defined (sinx+1)2+74≥2 (∵−1≤cos−1x≤1)⇒(sinx+1)2≥14⇒sinx+1≥12,    sinx+1≤−12             sinx≥−12,    sinx≤−32   (This is impossible)⇒x∈[0,  7π6]U[11π6,2π]   Hence (A) is correct