The correct option is A (π2,π]∪[3π2,2π]
The range of cosx is [−1,1] and thus the range of [cosx] becomes −1,0,1
Similarly, since the range of 1+sinx is [0,2]; the range of [1+sinx] becomes 0,1,2
Now, since the R.H.S. is 0, one of the terms on L.H.S will be 1 and the other -1 and the other possibility says both terms to be 0!
a) If [cosx]=0,cosx∈[0,1); also, [1+sinx]=0,1+sinx∈[0,1)
This is possible in the fourth quadrant.
b) If [cosx]=−1,cosx∈[−1,0) and also, [1+sinx]=1 so 1+sinx∈[1,2) i.e. sinx∈[0,1)
This is possible in the second quadrant.
Hence option A is correct!