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 :)