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

Question

37494 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T14:22:41

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

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 $(7pi)/4$ .

How do you solve sinx=-cosxsinx=-cosx in the interval 0<=x<=2pi0<=x<=2pi?