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Please solve the equation sinx+cosx-sin2x+cos2x-cos3x=1.

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sufjllaf | Student | (Level 1) eNoter

Posted March 22, 2012 at 9:45 AM via web

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Please solve the equation sinx+cosx-sin2x+cos2x-cos3x=1.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted March 22, 2012 at 2:24 PM (Answer #1)

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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-[cos^3x-3sin^2xcosx]=1`

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

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

(3) After dividing we are left with:

`1=2cosx=>cosx=1/2=>x=pi/3,(5pi)/3`

(4) So the complete set of solutions is:

`x=npi,n in ZZ`

`x=pi/6 + 2npi,n in ZZ`

`x=(5pi)/6+2npi,n in ZZ`

`x=pi/3+2npi,n in ZZ`

`x=(5pi)/3+2npi,n in ZZ`

The graph:

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