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



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Posted on (Answer #1)




Double angle identities

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




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

Move terms with just cos to right hand side:


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

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






(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:


(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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