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`sinx+cosx-sin2x+cos2x-cos3x=1`

Given

`sinx+cosx-2sinxcosx+cos^2x-sin^2x-cos3x=1`

Double angle identities

* `cos3x=cos(2x+x)=cos2xcosx-sin2xsinx`

`=(cos^2x-sin^2x)cosx-2sinxcosxsinx`

`=cos^3x-sin^2xcosx-2sin^2xcosx`

`=cos^3x-3sin^2xcosx`

So `sinx+cosx-2sinxcosx+cos^2x-sin^2x-=1`

Move terms with just cos khổng lồ right hand side:

`sinx-2sinxcosx-sin^2x+3sin^2xcosx=1-cosx-cos^2x+cos^3x`

On the righthand side, `1-cos^2x=sin^2x`

and `-cosx+cos^3x=cosx(-1+cos^2x)=cosx(-sin^2x)`

Then

`sinx-2sinxcosx-sin^2x+3sin^2xcosx=sin^2x-sin^2xcosx`

`sinx-2sinxcosx-2sin^2x+4sin^2xcosx=0`

`sinx-2sin^2x=2sinxcosx-4sin^2xcosx`

`sinx(1-2sinx)=2sinxcosx(1-2sinx)`

(1) We want lớn divide both sides by sinx; this...

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`sinx+cosx-sin2x+cos2x-cos3x=1`

Given

`sinx+cosx-2sinxcosx+cos^2x-sin^2x-cos3x=1`

Double angle identities

* `cos3x=cos(2x+x)=cos2xcosx-sin2xsinx`

`=(cos^2x-sin^2x)cosx-2sinxcosxsinx`

`=cos^3x-sin^2xcosx-2sin^2xcosx`

`=cos^3x-3sin^2xcosx`

So `sinx+cosx-2sinxcosx+cos^2x-sin^2x-=1`

Move terms with just cos lớn right hand side:

`sinx-2sinxcosx-sin^2x+3sin^2xcosx=1-cosx-cos^2x+cos^3x`

On the righthvà side, `1-cos^2x=sin^2x`

và `-cosx+cos^3x=cosx(-1+cos^2x)=cosx(-sin^2x)`

Then

`sinx-2sinxcosx-sin^2x+3sin^2xcosx=sin^2x-sin^2xcosx`

`sinx-2sinxcosx-2sin^2x+4sin^2xcosx=0`

`sinx-2sin^2x=2sinxcosx-4sin^2xcosx`

`sinx(1-2sinx)=2sinxcosx(1-2sinx)`

(1) We want to lớn divide both sides by sinx; this might be a solution so we must check when sinx=0. This is a solution to lớn the original problem so `x=npi,n in ZZ` (n an integer)

(2) We would also like to lớn divide by `1-2sinx` ; again khổng lồ avoid a lost root we kiểm tra the solution to `1-2sinx=0` in the original equation. Indeed, `sinx=50% =>x=pi/6,(5pi)/6` are both solutions.

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