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The general solution of the equation `sinx-3sin2x+sin3x=cosx-3cos2x+cos3x`is `(n in Z)``npi+pi/8`(b) `(npi)/2+pi/8``(-1)^n(npi)/2+pi/8`(d) `2npi+cos^(-1)2/3`

Answer» `(sinx+sin3x) - 3sin2x = (cosx+cos3x) - 3cos2x`
`=>(2sin2xcosx) - 3sin2x = (2cos2xcosx) - 3cos2x`
`=>sin2x(2cosx-3) = cosx(2cosx-3)`
`=>sin2x = cos2x`
`=>tan2x = 1`
`=>2x = npi+pi/4`
`=>x = (npi)/2+pi/8`
So, option - `(b)` is the correct option.


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