{ ((x + y = ((5π)/3))),((sin x = 2sin y)) :}


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Question Number 125359 by weltr last updated on 10/Dec/20

${\begin{cases} x + y = \frac{5 π}{3} \\ s i n x = 2 s i n y \end{cases}$
	Answered by Ar Brandon last updated on 10/Dec/20
	${ ((x+y=((5π)/3)),(...(i))),((sinx=2siny),(...(ii))) :} From (i), x+y=((5π)/3) ⇒ y=((5π)/3)−x substituting the above result in (ii) we have ; sinx=2sin(((5π)/3)−x)=2sin(π+(((2π)/3)−x))=2sin(x−((2π)/3)) {sin(π+x)=−sinx=sin(−x)} =2[sin(x)cos(((2π)/3))−cos(x)sin(((2π)/3))] =2[−(1/2)sinx−((√3)/2)cosx]=−sinx−(√3)cosx ⇒2sinx=−(√3)cosx ⇒ tanx=−((√3)/2) x=tan^(−1) (−((√3)/2)) , y=((5π)/3)−tan^(−1) (−((√3)/2))$
	${\begin{cases} x + y = \frac{5 π}{3} & . . . (i) \\ sinx = 2 siny & . . . (ii) \end{cases}$ $From (i), x + y = \frac{5 π}{3} \Rightarrow y = \frac{5 π}{3} - x$ $substituting the above result in (ii) we have;$ $sinx = 2 \sin (\frac{5 π}{3} - x) = 2 \sin (π + (\frac{2 π}{3} - x)) = 2 \sin (x - \frac{2 π}{3})$ ${s i n (π + x) = - s i n x = s i n (- x)}$ $= 2 [\sin (x) \cos (\frac{2 π}{3}) - \cos (x) \sin (\frac{2 π}{3})]$ $= 2 [- \frac{1}{2} sinx - \frac{\sqrt{3}}{2} cosx] = - sinx - \sqrt{3} cosx$ $\Rightarrow 2 sinx = - \sqrt{3} cosx \Rightarrow tanx = - \frac{\sqrt{3}}{2}$ $x = \tan^{- 1} (- \frac{\sqrt{3}}{2}), y = \frac{5 π}{3} - \tan^{- 1} (- \frac{\sqrt{3}}{2})$
	Commented by MJS_new last updated on 10/Dec/20

	$good path but you embezzle at least (10^{1000})!$ $solutions . . .$ $x = n π - \arctan \frac{\sqrt{3}}{2}$ $y = - n π + \frac{5 π}{3} + \arctan \frac{\sqrt{3}}{2}$
	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Dwaipayan Shikari last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Dwaipayan Shikari last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Ar Brandon last updated on 10/Dec/20
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	Commented by Dwaipayan Shikari last updated on 10/Dec/20
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	Answered by Dwaipayan Shikari last updated on 10/Dec/20

	$\frac{s i n x + s i n y}{s i n x - c o s y} = 3 \Rightarrow \frac{t a n (\frac{x + y}{2})}{t a n (\frac{x - y}{2})} = 3 \Rightarrow t a n (\frac{x - y}{2}) = - \frac{1}{3 \sqrt{3}}$ $x - y = 2 k π - {t a n}^{- 1} \frac{1}{3 \sqrt{3}}$ $x = (k + \frac{5}{6}) π - \frac{1}{2} {t a n}^{- 1} \frac{1}{3 \sqrt{3}}, y = (\frac{5}{6} - k) π + \frac{1}{2} {t a n}^{- 1} \frac{1}{3 \sqrt{3}}$
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