The correct option is C [sin 1, 1]
[sin−1x]=−2,−1,0,1[∵−π2≤sin−1x≤π2]0≤cos−1x≤π⇒[cos−1x]=0,1,2,3.
∴ Possible values for [sin−1x] and [cos−1x] satisfying [sin−1x]>[cos−1x] are [sin−1x]=1 and [cos−1x]=0
i.e., 1≤sin−1x≤π2,0≤cos−1x<1⇒sin1≤x≤1,cos1≤x≤1.
Since 1>π4,sin1>cos1
Hence, required solution is x∈[sin1,1]