Question

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

Answer 1

`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`

Answer 2

`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

Answer 3

`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 :)