The general solution of the equation `sin^(100)x-cos^(100)x=1` is `2npi+pi/3,n in I` (b) n`pi+pi/2,n in I` `npi+pi/4,n in I` (d) `2npi=pi/3,n in I`

Options

A. `2npi+pi/3, n in I`
B. `n pi+pi/2, n in I`
C. `n pi+pi/4, n in I`
D. `2n pi-pi/3, n in I`

Answer

B

Explanation

We have `sin^(100)x-cos^(100)x=1`
or `sin^(100)x=1+cos^(100)x`
Since L.H.S. `le 1` and R.H.S. `ge 1`, L.H.S. = R.H.S.=1.
Then, `x=n pi+pi/2, n in I`