What is the general solution for sin(x)+ cos(x)=-1?

As `sinx+cosx=-1`, we have

`sqrt2(sinx*1/sqrt2+cosx*1/sqrt2)=-1`

or `sinxcos(pi/4)+cosxsin(pi/4)=-1/sqrt2`

or `sin(x+pi/4)=sin(-pi/4)`

Now general solution of `sinx=sinalpha` is `x=npi+-(-1)^nalpha`

Hence, general solution is

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

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

or `x=npi-(-1)^n(pi/4)-pi/4`

`sin(x)+cos(x)=-1`

`=>sin(x)=-1-cos(x)` equation-1

We have an identity

`sin^2(x)+cos^2(x)=1`

Use this to find the value of `sin(x)`

`=> sin^2(x)=1-cos^2(x)`

`=> sin(x)=+-sqrt(1-cos^2(x))`

We got two values for `sin(x)`

`+sqrt(1-cos^2(x)) "and" -sqrt(1-cos^2(x))`

Put them one by one in equation-1.

`=> +sqrt(1-cos^2(x))=-1-cos(x)`

Squaring both sides

`=> 1-cos^2(x)=1+cos^2(x)+2cos(x)`

`=> 0=2cos^2(x)+2cos(x)`

Divide by two both sides

`=> 0=cos^2(x)+cos(x)`

`=> 0=cos(x)(cos(x)+1)`

It gives `cos(x)=0`

We get `sin(x)=-1`

The solution for this is

`x = 3/2 pi + 2pin`

Here , `pi = 180^@` and n is any integer.

Now , we also get `cos(x)+1=0`

`=> cos(x)=-1`

It gives `sin(x)=0` according to the given equation.

The solution for this is

`x= pi + 2pin`

If you put `sin(x)=-sqrt(1-cos^2(x))` you get the same results.

Hope it helps :)