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

2 Answers
Ratnaker Mehta
May 24, 2018

#"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#.

Abhishek K.
May 24, 2018

#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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