We have `[cos x]+|sin x|=1`
Now, `-1 le cos x le 1`
`:.` Possible values of `[cos x]` are `0, 1, -1`
Case I :
`[cos x]=-1` or `cos x in [-1, 0)`
`:. -1 +|sin x|=1`
`rArr |sin x|=2` (not possible)
Case II :
`[cos x]=0` or `cos x in [0, 1)`
`:. |sin x|=1`
`rArr x=(3pi)/2, (5pi)/2`
Case III :
`[cos x]=1` or `cos x=1`
`:. |sin x|=0`
`:. x=2pi`
Hence, there are three solutions.