Question

How do you solve `sinx=-cosx` in the interval `0<=x<=2pi`?

Answer 1

`x=(3pi)/4` or `(7pi)/4`

Explanation:

As `sinx=-cosx`, we have

`sinx+cosx=0`

or `sinx/sqrt2+cosx/sqrt2=0`

or `sinxcos(pi/4)+cosxsin(pi/4)=0`

or `sin(x+pi/4)=0`=sin0#

or `sin(x+pi/4)=sin0` or `sinpi` or `sin2pi`

Hence possible values of `x` in the interval `0<=x<=2pi` is

`x=pi-pi/4=(3pi)/4` or `x=2pi-pi/4=(7pi)/4`

Alternatively `sinx=-cosx=>tanx=-1`

i.e. `x=(3pi)/4` or `(7pi)/4`

An easier way could be that as `sinx=-cosx`

`sinx/cosx=-1` or `tanx=tan(-pi/4)`

and as tan ratio has a cylce of `pi`

`x={-pi/4,(3pi)/4,(7pi)/4,......}` and possible values of `x` in the interval `0<=x<=2pi` are `(3pi)/4` and `(7pi)/4`.