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Question #d4016

3 Answers

Dec 15, 2016

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

Explanation:

Remember that $cos (x) = \pm \sqrt{1 - {sin}^{2} (x)}$ $cos(x)=+-sqrt(1-sin^2(x))$

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

$XXX \to sin (x) - 7 (\pm \sqrt{1 - {sin}^{2} (x)}) = 7$ $color(white)("XXX")rarr sin(x)-7(+-sqrt(1-sin^2(x)))=7$

$XXX \to sin (x) - 7 = 7 (\pm \sqrt{1 - {sin}^{2} (x)})$ $color(white)("XXX")rarr sin(x)-7 =7 (+-sqrt(1-sin^2(x)))$

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

$XXX \to {sin}^{2} (x) - 14 sin (x) = 49 {sin}^{2} (x)$ $color(white)("XXX")rarr sin^2(x)-14sin(x) =49sin^2(x)$

$XXX \to 50 {sin}^{2} (x) - 14 sin (x) = 0$ $color(white)("XXX")rarr 50sin^2(x)-14sin(x) =0$

$XXX \to sin (x) (50 sin (x) - 14) = 0$ $color(white)("XXX")rarr sin(x)(50sin(x)-14)=0$

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

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

Dec 15, 2016

$sinx-7cosx=7$

$=>sinx-7cosx-7=0$

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

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

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

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

$=>x=2npi+2alpha$

Dec 15, 2016

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

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