Question

Question #d4016

Answer 1

Within the domain `x in[0,2pi)`
`sin(x)-7cos(x)=7color(white)("XX")rarrcolor(white)("XX")x = pi or x~~2.8578 ("radians")`

Explanation:

Remember that `cos(x)=+-sqrt(1-sin^2(x))`

Therefore
`color(white)("XXX")sin(x)-7cos(x)=7`

`color(white)("XXX")rarr sin(x)-7(+-sqrt(1-sin^2(x)))=7`

`color(white)("XXX")rarr sin(x)-7 =7 (+-sqrt(1-sin^2(x)))`

squaring both sides (caution: extraneous solutions may be introduced at this point)
`color(white)("XXX")rarr sin^2(x)-14sin(x)+49=49 * ( 1-sin^2(x))`

`color(white)("XXX")rarr sin^2(x)-14sin(x) =49sin^2(x)`

`color(white)("XXX")rarr 50sin^2(x)-14sin(x) =0`

`color(white)("XXX")rarr sin(x)(50sin(x)-14)=0`

`color(white)("XXX")rarr`
#color(white)("XXXXX"){: (sin(x)=0,color(white)("XX")orcolor(white)("XX"),sin(x)=14/50), (x=0 or x=pi,,x=arcsin(14/50) or x=pi-arcsin(14/50)), (,,x~~0.28379 or x~~2.85780 ("radians")) :}#

Testing these four possible solutions against the original equation
shows that `x=0 and x~~0.28379` are extraneous.

Answer 2

`sinx-7cosx=7`

`=>sinx-7cosx-7=0`

`=>2sin(x/2)cos(x/2)-7(cosx+1)=0`

`=>2sin(x/2)cos(x/2)-7*2cos^2(x/2)=0`

`=>2cos(x/2)(sin(x/2)-7cos(x/2))=0`

So `cos(x/2)=0`

`x/2=(2n+1)pi/2" where "n in ZZ`

`=>x=(2n+1)pi`

Again

`sin(x/2)=7cos(x/2)`

`=>tan(x/2)=7=tanalpha`

`" where "alpha=tan^-1(7) ~~1.43"radian"`

`"and when "cos(x/2)!=0`

So `x/2=npi+alpha " where "n in ZZ`

`=>x=2npi+2alpha`

Answer 3

The answer is `x=2.83rad`

Explanation:

We can use

`Rsin(x-alpha)=R(sinxcosalpha-cosxsinalpha)`

Comparing this to `sinx-7cosx`

We see that

`Rcosalpha=1` and `Rsinalpha=7`

`R^2cos^2alpha+R^2sin^2alpha=1+49=50`

`R=sqrt50`

`cosalpha=1/sqrt50` , `=>`, `alpha=1.43rad`

`sinalpha=7/sqrt50=1.43rad`

Therefore

`Rsin(x-alpha)=7`

`sin(x-1.43)=7/sqrt50`

`x-1.43=1.43`

`x=2.86`

Q.E.D