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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`

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`sin^100x - cos^100x = 1`
`=>sin^100x = 1+cos^100x`
Now, maximum value of `sin^100x` can be `1`.
So, above equation will satisfy only when `cos^100x =0.`
`:. sin^100x = 1 and cos^100x = 0` is the only solution for this equation.
Now, `sin^100x = 1`
`=>sin^2x = 1`
`=>sinx = +-1`
...
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