How do you solve $sin x = - cos x$ $sinx=-cosx$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2018-03-07T14:22:41

$x = \frac{3 π}{4}$ $x=(3pi)/4$ or $\frac{7 π}{4}$ $(7pi)/4$

As $sin x = - cos x$ $sinx=-cosx$ , we have

$sin x + cos x = 0$ $sinx+cosx=0$

or $\frac{sin x}{\sqrt{2}} + \frac{cos x}{\sqrt{2}} = 0$ $sinx/sqrt2+cosx/sqrt2=0$

or $sin x cos (\frac{π}{4}) + cos x sin (\frac{π}{4}) = 0$ $sinxcos(pi/4)+cosxsin(pi/4)=0$

or $sin (x + \frac{π}{4}) = 0$ $sin(x+pi/4)=0$ =sin0#

or $sin (x + \frac{π}{4}) = sin 0$ $sin(x+pi/4)=sin0$ or $sin π$ $sinpi$ or $sin 2 π$ $sin2pi$

Hence possible values of $x$ $x$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ is

$x = π - \frac{π}{4} = \frac{3 π}{4}$ $x=pi-pi/4=(3pi)/4$ or $x = 2 π - \frac{π}{4} = \frac{7 π}{4}$ $x=2pi-pi/4=(7pi)/4$

Alternatively $sin x = - cos x \Rightarrow tan x = - 1$ $sinx=-cosx=>tanx=-1$

i.e. $x = \frac{3 π}{4}$ $x=(3pi)/4$ or $\frac{7 π}{4}$ $(7pi)/4$

An easier way could be that as $sin x = - cos x$ $sinx=-cosx$

$\frac{sin x}{cos x} = - 1$ $sinx/cosx=-1$ or $tan x = tan (- \frac{π}{4})$ $tanx=tan(-pi/4)$

and as tan ratio has a cylce of $π$ $pi$

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

How do you solve sinx=-cosxsinx=−cosxsinx=-cosx in the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?