Question

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

Answer 1

`"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

`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`