The general solution of the equation sin100⁡x−cos100⁡x=1 is

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2nπ+π3,n∈I

nπ+π2,n∈I

nπ+π4,n∈I

2nπ−π3,n∈I

answer is B.

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Detailed Solution

We have sin100⁡x−cos100⁡x=1or sin100⁡x=1+cos100⁡xSince L.H.S. ≤ 1 and R.H.S. ≥ 1, L.H.S. = R.H.S. = 1then , x=nπ+π2,n∈I

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