{ ((4xy=1)),((4(√(1−x^2 )) ( y−(√(1−y^2 )) )=1)) :} Resolver elsistema en R


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Question Number 72488 by aliesam last updated on 29/Oct/19
${ ((4xy=1)),((4(√(1−x^2 )) ( y−(√(1−y^2 )) )=1)) :} Resolver elsistema en R$
${\begin{cases} 4 x y = 1 \\ 4 \sqrt{1 - x^{2}} (y - \sqrt{1 - y^{2}}) = 1 \end{cases}$ $R e s o l v e r e l s i s t e m a e n R$
	Answered by behi83417@gmail.com last updated on 30/Oct/19
	$x=cost,y=cosr ⇒ { ((4sint(cosr−sinr)=1)),((4cost.cosr=1)) :} ⇒(1/((cosr−sinr)^2 ))+(1/(cos^2 r))=4⇒ ⇒cos^2 r+(cosr−sinr)^2 =4cos^2 r.(cosr−sinr)^2 ((1+cos2r)/2)+1−sin2r=2(1+cos2r)(1−sin2r) ⇒1+cos2r+2−2sin2r= 4(1−sin2r+cos2r−sin2rcos2r) ⇒3+cos2r−2sin2r= 4−4sin2r+4cos2r−4sin2rcos2r ⇒3cos2r−2sin2r−4sin2rcos2r+1=0 [let:c=cos2r,s=sin2r]⇒ 3c−2s−4sc+1=0 9c^2 +4s^2 −12sc=16s^2 c^2 −8sc+1 9c^2 +4s^2 −16s^2 c^2 =4sc+1 5c^2 −16c^2 (1−c^2 )+3=4sc ⇒16c^4 −11c^2 +3=4sc ⇒256c^8 +121c^4 +9−352c^6 +96c^4 −66c^2 = =16c^2 −16c^4 ⇒256c^8 −352c^6 +233c^4 −82c^2 +9=0 ⇒c=±0.42,±0.76 ⇒cos2r=±0.42,±0.76 ⇒ { ((cos2r=0.42⇒y=cosr=0.843)),((⇒x=(1/(4y))=(1/(4×0.843))=0.211)) :} ⇒ { ((cos2r=−0.42⇒y=cosr=0.54)),((x=(1/(4y))=(1/(4×0.211))=0.053)) :} ⇒ { ((cos2r=0.76⇒y=cosr=0.938)),((x=(1/(4y))=(1/(4×0.938))=0.235)) :} ⇒ { ((cos2r=−0.76⇒y=cosr=0.346)),((x=(1/(4y))=(1/(4×0.346))=0.087)) :}$
	$x = cost, y = cosr$ $\Rightarrow {\begin{cases} 4 sint (cosr - sinr) = 1 \\ 4 cost . cosr = 1 \end{cases}$ $\Rightarrow \frac{1}{{(cosr - sinr)}^{2}} + \frac{1}{\cos^{2} r} = 4 \Rightarrow$ $\Rightarrow \cos^{2} r + {(cosr - sinr)}^{2} = {4 \cos}^{2} r . {(cosr - sinr)}^{2}$ $\frac{1 + \cos 2 r}{2} + 1 - \sin 2 r = 2 (1 + \cos 2 r) (1 - \sin 2 r)$ $\Rightarrow 1 + \cos 2 r + 2 - 2 \sin 2 r =$ $4 (1 - \sin 2 r + \cos 2 r - \sin 2 rcos 2 r)$ $\Rightarrow 3 + \cos 2 r - 2 \sin 2 r =$ $4 - 4 \sin 2 r + 4 \cos 2 r - 4 \sin 2 rcos 2 r$ $\Rightarrow 3 \cos 2 r - 2 \sin 2 r - 4 \sin 2 rcos 2 r + 1 = 0$ $[let : c = \cos 2 r, s = \sin 2 r] \Rightarrow$ $3 c - 2 s - 4 sc + 1 = 0$ ${9 c}^{2} + {4 s}^{2} - 12 sc = {16 s}^{2} c^{2} - 8 sc + 1$ ${9 c}^{2} + {4 s}^{2} - {16 s}^{2} c^{2} = 4 sc + 1$ ${5 c}^{2} - {16 c}^{2} (1 - c^{2}) + 3 = 4 sc$ $\Rightarrow {16 c}^{4} - {11 c}^{2} + 3 = 4 sc$ $\Rightarrow {256 c}^{8} + {121 c}^{4} + 9 - {352 c}^{6} + {96 c}^{4} - {66 c}^{2} =$ $= {16 c}^{2} - {16 c}^{4}$ $\Rightarrow {256 c}^{8} - {352 c}^{6} + {233 c}^{4} - {82 c}^{2} + 9 = 0$ $\Rightarrow c = \pm 0.42, \pm 0.76$ $\Rightarrow \cos 2 r = \pm 0.42, \pm 0.76$ $\Rightarrow {\begin{cases} \cos 2 r = 0.42 \Rightarrow y = cosr = 0.843 \\ \Rightarrow x = \frac{1}{4 y} = \frac{1}{4 \times 0.843} = 0.211 \end{cases}$ $\Rightarrow {\begin{cases} \cos 2 r = - 0.42 \Rightarrow y = cosr = 0.54 \\ x = \frac{1}{4 y} = \frac{1}{4 \times 0.211} = 0.053 \end{cases}$ $\Rightarrow {\begin{cases} \cos 2 r = 0.76 \Rightarrow y = cosr = 0.938 \\ x = \frac{1}{4 y} = \frac{1}{4 \times 0.938} = 0.235 \end{cases}$ $\Rightarrow {\begin{cases} \cos 2 r = - 0.76 \Rightarrow y = cosr = 0.346 \\ x = \frac{1}{4 y} = \frac{1}{4 \times 0.346} = 0.087 \end{cases}$
	Commented by aliesam last updated on 29/Oct/19

	$g o d b l e s s y o u$
	Answered by MJS last updated on 29/Oct/19

	$(1) \Rightarrow x = \frac{1}{4 y}$ $\Rightarrow$ $(2) \frac{\sqrt{16 y^{2} - 1} (y - \sqrt{1 - y^{2}})}{∣ y ∣} = 1$ $\sqrt{16 y^{2} - 1} (y - \sqrt{1 - y^{2}}) - ∣ y ∣= 0$ $\Rightarrow - 1 ⩽ y ⩽ - \frac{1}{4} \lor \frac{1}{4} ⩽ y ⩽ 1$ $we can only approximate (it leads to a$ $biquartic in y \Rightarrow quartic in \sqrt{y} which has$ $no ‘ ‘ {nice}^{″} solution)$ $the only solution I get is$ $x \approx . 302667; y \approx . 825989$
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