`sintheta-costheta = 1`

Find the general solution.

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Solve `sintheta-costheta=1` :

Square both sides (be aware that you might introduce extraneous solutions):

`sin^2theta-2sinthetacostheta+cos^2theta=1`

Use the Pythagorean identity and the double angle identity:

`1-sin2theta=1`

`sin2theta=0`

`2theta=0,pi`

`theta=0,pi/2,pi,(3pi)/2` on `[0,2pi)`

Checking for extraneous solutions we find that only `theta=pi/2,theta=pi` are solutions. (`theta=0` gives `sin0-cos0=-1` and `theta=(3pi)/2` gives `sin(3pi)/2-cos(3pi)/2=-1` )

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The general solution is `theta=pi/2 +2npi,theta=pi+2npi` ,`n in ZZ`

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`sin(theta)-cos(theta)=1`

`cos(pi/2-theta)-cos(theta)=1`

`2sin((((pi/2-theta)+theta))/2)sin((theta-(pi/2-theta))/2)=1`

`2sin(pi/4)sin(theta-pi/4)=1`

`2(1/sqrt(2))sin(theta-pi/4)=1`

`sin(theta-pi/4)=1/sqrt(2)`

`sin(theta-pi/4)=sin(pi/4)`

`theta-pi/4=npi+(-1)^n(pi/4)`

`theta=npi+pi/4+(-1)^n(pi/4)`

`theta=npi+(1+(-1)^n)(pi/4)`

**where n is an integer.It is general solution.**

We have applied two formulas

`cos(c)-cos(D)=2sin((C+D)/2)sin((D-C)/2)`

`cos(90-theta)=sin(theta)`

`But`

`sin(90+-theta)=cos(theta)`

So I have not cosider this .For cos(theta) there are two different value ,we need to choose unique .So I changed sin(theta).

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