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

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

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Answer 1 · 2018-05-24T14:30:31

$"The Solution Set"={2kpi}uu{kpi+pi/4}, k in ZZ$ .

Explanation:

Given that, $sinx-cosx-tanx=-1$ .

$:. sinx-cosx-sinx/cosx+1=0$ .

$:. (sinx-cosx)-(sinx/cosx-1)=0$ .

$:. (sinx-cosx)-(sinx-cosx)/cosx=0$ .

$:. (sinx-cosx)cosx-(sinx-cosx)=0$ .

$:. (sinx-cosx)(cosx-1)=0$ .

$:. sinx=cosx or cosx=1$ .

$" Case 1 : "sinx=cosx$ .

Observe that $cosx!=0, because," if otherwise; "tanx" becomes"$

undefined.

Hence, dividing by $cosx!=0, sinx/cosx=1, or, tanx=1$ .

$:. tanx=tan(pi/4)$ .

$:. x=kpi+pi/4, k in ZZ," in this case"$ .

$" Case 2 : "cosx=1$ .

$"In this case, "cosx=1=cos0, :. x=2kpi+-0, k in ZZ$ .

Altogether, we have,

$"The Solution Set"={2kpi}uu{kpi+pi/4}, k in ZZ$ .

Answer 2 · 2018-05-24T14:37:02

$rarrx=2npi,npi+pi/4$ where $n in ZZ$

Explanation:

$rarrsinx-cosx-tanx=-1$

$rarrsinx-cosx-sinx/cosx+1=0$

$rarr(sinx*cosx-cos^2x-sinx+cosx)/cosx=0$

$rarrsinx*cosx-sinx-cos^2x+cosx=0$

$rarrsinx(cosx-1)-cosx(cosx-1)=0$

$rarr(cosx-1)(sinx-cosx)=0$

When $rarrcosx-1=0$

$rarrcosx=cos0$

$rarrx=2npi+-0=2npi$ where $n in ZZ$

When $rarrsinx-cosx=0$

$rarrcos(90-x)-cosx=0$

$rarr2sin((90-x+x)/2)*sin((x-90+x)/2)=0$

$rarrsin(x-pi/4)=0$ As $sin(pi/4)!=0$

$rarrx-pi/4=npi$

$rarrx=npi+pi/4$ where $n in ZZ$

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

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