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4xy-1-4-1-x-2-y-1-y-2-1-Resolver-elsistema-en-R-




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
$$\begin{cases}{\mathrm{4}{xy}=\mathrm{1}}\\{\mathrm{4}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\left(\:{y}−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:\right)=\mathrm{1}}\end{cases} \\ $$$$ \\ $$$${Resolver}\:{elsistema}\:{en}\:{R} \\ $$
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)) :}
$$\mathrm{x}=\mathrm{cost},\mathrm{y}=\mathrm{cosr} \\ $$$$\Rightarrow\begin{cases}{\mathrm{4sint}\left(\mathrm{cosr}−\mathrm{sinr}\right)=\mathrm{1}}\\{\mathrm{4cost}.\mathrm{cosr}=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{cosr}−\mathrm{sinr}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{r}}=\mathrm{4}\Rightarrow \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \mathrm{r}+\left(\mathrm{cosr}−\mathrm{sinr}\right)^{\mathrm{2}} =\mathrm{4cos}^{\mathrm{2}} \mathrm{r}.\left(\mathrm{cosr}−\mathrm{sinr}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}+\mathrm{cos2r}}{\mathrm{2}}+\mathrm{1}−\mathrm{sin2r}=\mathrm{2}\left(\mathrm{1}+\mathrm{cos2r}\right)\left(\mathrm{1}−\mathrm{sin2r}\right) \\ $$$$\Rightarrow\mathrm{1}+\mathrm{cos2r}+\mathrm{2}−\mathrm{2sin2r}= \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{sin2r}+\mathrm{cos2r}−\mathrm{sin2rcos2r}\right) \\ $$$$\Rightarrow\mathrm{3}+\mathrm{cos2r}−\mathrm{2sin2r}= \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}−\mathrm{4sin2r}+\mathrm{4cos2r}−\mathrm{4sin2rcos2r} \\ $$$$\Rightarrow\mathrm{3cos2r}−\mathrm{2sin2r}−\mathrm{4sin2rcos2r}+\mathrm{1}=\mathrm{0} \\ $$$$\left[\mathrm{let}:\mathrm{c}=\mathrm{cos2r},\mathrm{s}=\mathrm{sin2r}\right]\Rightarrow \\ $$$$\mathrm{3c}−\mathrm{2s}−\mathrm{4sc}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{9c}^{\mathrm{2}} +\mathrm{4s}^{\mathrm{2}} −\mathrm{12sc}=\mathrm{16s}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} −\mathrm{8sc}+\mathrm{1} \\ $$$$\mathrm{9c}^{\mathrm{2}} +\mathrm{4s}^{\mathrm{2}} −\mathrm{16s}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} =\mathrm{4sc}+\mathrm{1} \\ $$$$\mathrm{5c}^{\mathrm{2}} −\mathrm{16c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{c}^{\mathrm{2}} \right)+\mathrm{3}=\mathrm{4sc} \\ $$$$\Rightarrow\mathrm{16c}^{\mathrm{4}} −\mathrm{11c}^{\mathrm{2}} +\mathrm{3}=\mathrm{4sc} \\ $$$$\Rightarrow\mathrm{256c}^{\mathrm{8}} +\mathrm{121c}^{\mathrm{4}} +\mathrm{9}−\mathrm{352c}^{\mathrm{6}} +\mathrm{96c}^{\mathrm{4}} −\mathrm{66c}^{\mathrm{2}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{16c}^{\mathrm{2}} −\mathrm{16c}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{256c}^{\mathrm{8}} −\mathrm{352c}^{\mathrm{6}} +\mathrm{233c}^{\mathrm{4}} −\mathrm{82c}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{c}=\pm\mathrm{0}.\mathrm{42},\pm\mathrm{0}.\mathrm{76} \\ $$$$\Rightarrow\mathrm{cos2r}=\pm\mathrm{0}.\mathrm{42},\pm\mathrm{0}.\mathrm{76} \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos2r}=\mathrm{0}.\mathrm{42}\Rightarrow\mathrm{y}=\mathrm{cosr}=\mathrm{0}.\mathrm{843}}\\{\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{4y}}=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{0}.\mathrm{843}}=\mathrm{0}.\mathrm{211}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos2r}=−\mathrm{0}.\mathrm{42}\Rightarrow\mathrm{y}=\mathrm{cosr}=\mathrm{0}.\mathrm{54}}\\{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{4y}}=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{0}.\mathrm{211}}=\mathrm{0}.\mathrm{053}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos2r}=\mathrm{0}.\mathrm{76}\Rightarrow\mathrm{y}=\mathrm{cosr}=\mathrm{0}.\mathrm{938}}\\{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{4y}}=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{0}.\mathrm{938}}=\mathrm{0}.\mathrm{235}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos2r}=−\mathrm{0}.\mathrm{76}\Rightarrow\mathrm{y}=\mathrm{cosr}=\mathrm{0}.\mathrm{346}}\\{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{4y}}=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{0}.\mathrm{346}}=\mathrm{0}.\mathrm{087}}\end{cases} \\ $$
Commented by aliesam last updated on 29/Oct/19
god bless you
$${god}\:{bless}\:{you} \\ $$
Answered by MJS last updated on 29/Oct/19
(1)  ⇒ x=(1/(4y))  ⇒  (2)  (((√(16y^2 −1))(y−(√(1−y^2 ))))/(∣y∣))=1  (√(16y^2 −1))(y−(√(1−y^2 )))−∣y∣=0  ⇒ −1≤y≤−(1/4) ∨ (1/4)≤y≤1  we can only approximate (it leads to a  biquartic in y ⇒ quartic in (√y) which has  no “nice” solution)  the only solution I get is  x≈.302667; y≈.825989
$$\left(\mathrm{1}\right)\:\:\Rightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{4}{y}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:\:\frac{\sqrt{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{1}}\left({y}−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\right)}{\mid{y}\mid}=\mathrm{1} \\ $$$$\sqrt{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{1}}\left({y}−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\right)−\mid{y}\mid=\mathrm{0} \\ $$$$\Rightarrow\:−\mathrm{1}\leqslant{y}\leqslant−\frac{\mathrm{1}}{\mathrm{4}}\:\vee\:\frac{\mathrm{1}}{\mathrm{4}}\leqslant{y}\leqslant\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}\:\left(\mathrm{it}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\right. \\ $$$$\mathrm{biquartic}\:\mathrm{in}\:{y}\:\Rightarrow\:\mathrm{quartic}\:\mathrm{in}\:\sqrt{{y}}\:\mathrm{which}\:\mathrm{has} \\ $$$$\left.\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution}\right) \\ $$$$\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{I}\:\mathrm{get}\:\mathrm{is} \\ $$$${x}\approx.\mathrm{302667};\:{y}\approx.\mathrm{825989} \\ $$

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