# How do you solve "Sin(πx) = Cos(πx) - 1"?

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Given `sin(pix)=cos(pix)-1` :

`sin^2(pix)=cos^2(pix)-2cos(pix)+1` Square both sides

`1-cos^2(pix)=cos^2(pix)-2cos(pix)+1` Pythagorean identity

`2cos^2(pix)-2cos(pix)=0`

`2cos(pix)(cos(pix)-1)=0`

Either `2cos(pix)=0` or `cos(pix)-1=0`

(1) `cos(pix)-1=0 => cos(pix)=1 => pix=cos^(-1)(1)`

Then `x=2n` for `n in ZZ` (n an integer)

(2) `2cos(pix)=0=>pix=cos^(-1)(0)`

(i) `pix=pi/2 => x=1/2 +- 2n, n in ZZ` However, this is an extraneous solution as it is not a solution to the original problem. (`1!=-1` ) This extraneous solution was created when we squared both sides.

(ii) `pix=-pi/2=>x=2n-1/2, n in ZZ`

**(3) Thus the solution is `x=2n` or `x=2n-1/2` for `n in ZZ` **