How do you solve sin(x)+cos(x)=-1?

3 Answers
Chirag Mehta · Shwetank Mauria
Mar 8, 2018

#x=2npi+pi# or #x=2npi-pi/2#

Explanation:

Given: #sinx+cosx=-1#

Square both sides

#(sinx+cosx)^2=(-1)^2#

#sin^2x+cos^2x+2sinxcosx=1#

#1+2sinxcosx=1#

#2sinxcosx=0#

Either #sinx=0" or "cosx=0#

If #sinx=0#
#cosx=-1#

and #x=2npi+pi#

If #cosx=0#
#sinx=-1#

and #x=2npi-pi/2#

Hriman
Mar 8, 2018

#x= pi+2pin, n∈Z#
#x= (3pi)/2+2pin, n∈Z#

Explanation:

  1. Square both sides
    #(sin(x)+cos(x))^2=(-1)^2#
    #sin(x)^2+cos^2x+2sinxcosx=1#
    #1+2sinxcosx=1#
    #2sinxcosx=0#
    #2sinx(cosx)=0#
    #sinx=0#
    #cosx=0#
    #x= pi+2pin, n∈Z#
    #x= (3pi)/2+2pin, n∈Z#
    #x=pi/2+2pin, n∈Z# EXTRANEOUS
    #x=0+2pin, n∈Z# EXTRANEOUS
Ujjwal
Mar 8, 2018

#x = 3/2 pi + 2pin " or " pi + 2pin#

Explanation:

#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 :)

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