Pls fast: Example 29. The domain of the function. f x = sin - 1 sin x - cos - 1 cos x i n 0 , 2 π i s . a 0 , π 2 ∪ 3 π 2 , 2 π b π , 2 π c 0 , π π 2 d 0 , 2 π - π 2 , 3 π 2 Share with your friends Share 0 Aarushi Mishra answered this fx=sin-1sin x-cos-1cos xCase 1: x∈0,π2 In interval 0,π2, sin-1sin x=x and cos-1cos x=x fx=x-xSince x≥0, so x=xfx=x-x=0Therefore 0,π2 is in domain of fxCase 2: x∈(π2,π] In interval (π2,π], sin-1sin x=sin-1sin π-x=π-x and cos-1cos x=x fx=π-x-xSince x∈(π2,π], so π-x≥0, therefore π-x=π-xfx=π-x-x=π-2xquantity inside square root should be non negative⇒π-2x≥02x≤πx≤π2But x∈(π2,π]Therefore no solution existsTherefore (π2,π] is not in domain of fxCase 3: x∈π,3π2In interval π,3π2, sin-1sin x=sin-1sin π-x=π-x and cos-1cos x=cos-1cos 2π-x=2π-x fx=π-x-2π-xSince x∈π,3π2, so π-x<0, therefore π-x=-π-x fx=-π-x-2π-x =-π+x-2π+x=-3π+2xquantity inside square root should be non negative⇒-3π+2x≥02x≥3πx≥3π2But x∈(π,3π2)Therefore no solution existsTherefore π,3π2 is not in domain of fxCase 4: x∈3π2,2πIn interval 3π2,2π, sin-1sin x=sin-1-sin 2π-x=sin-1sin x-2π=x-2π and cos-1cos x=cos-1cos 2π-x=2π-x fx=x-2π -2π-xSince x∈3π2,2π, so x-2π<0, therefore x-2π=-x-2π fx=-x-2π-2π-x =-x+2π-2π+x=0Therefore 3π2,2π is in domain of fxHenceDomain=0,π2∪3π2,2π 0 View Full Answer