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Question Number 166468 by mathlove last updated on 20/Feb/22

1+3(√2)x−18x^2 =6x(√(1−9x^2 )) solve in R

$$\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}{x}−\mathrm{18}{x}^{\mathrm{2}} =\mathrm{6}{x}\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} } \\ $$$${solve}\:\:{in}\:{R} \\ $$

Answered by MJS_new last updated on 20/Feb/22

squaring & transforming ⇒ x^4 −((√2)/6)x^3 −(1/(12))x^2 +((√2)/(108))x+(1/(648))=0 (x^2 −(1/(18)))(x^2 −((√2)/6)x−(1/(36)))=0 of the 4 solutions only 3 fit the given equation x=±((√2)/6)∨x=(((√6)+(√2))/(12))

$$\mathrm{squaring}\:\&\:\mathrm{transforming}\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{2}} +\frac{\sqrt{\mathrm{2}}}{\mathrm{108}}{x}+\frac{\mathrm{1}}{\mathrm{648}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{18}}\right)\left({x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}{x}−\frac{\mathrm{1}}{\mathrm{36}}\right)=\mathrm{0} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{only}\:\mathrm{3}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}\vee{x}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$

Answered by mr W last updated on 20/Feb/22

let 3x=sin θ 1+(√2) sin θ−2 sin^2 θ=2 sin θ cos θ cos 2θ+(√2) sin θ=sin 2θ sin θ=(1/( (√2)))(sin 2θ−cos 2θ) sin θ=sin (2θ−(π/4)) 2θ−(π/4)=2kπ+θ ⇒θ=2kπ+(π/4) 2θ−(π/4)=(2k+1)π−θ ⇒θ=((2kπ)/3)+((5π)/(12)) ⇒x_1 =(1/3) sin (π/4)=((√2)/6) ⇒x_2 =(1/3) sin ((5π)/(12))=(((√6)+(√2))/(12)) ⇒x_3 =−(1/3) sin ((π/4))=−((√2)/6) ⇒x=(1/3) sin (((13π)/(12))) (rejected, since cos ((13π)/(12))<0)

$${let}\:\mathrm{3}{x}=\mathrm{sin}\:\theta \\ $$$$\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\mathrm{2}\theta+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta=\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{sin}\:\mathrm{2}\theta−\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\mathrm{sin}\:\theta=\mathrm{sin}\:\left(\mathrm{2}\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{2}\theta−\frac{\pi}{\mathrm{4}}=\mathrm{2}{k}\pi+\theta\:\Rightarrow\theta=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{2}\theta−\frac{\pi}{\mathrm{4}}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi−\theta\:\Rightarrow\theta=\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{4}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$$\Rightarrow{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{12}}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$$$\Rightarrow{x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)=−\frac{\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{13}\pi}{\mathrm{12}}\right)\:\left({rejected},\:{since}\:\mathrm{cos}\:\frac{\mathrm{13}\pi}{\mathrm{12}}<\mathrm{0}\right) \\ $$

Commented by peter frank last updated on 20/Feb/22

Thank you both

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{both} \\ $$

Commented by MaxiMaths last updated on 20/Feb/22

ok

$$\mathrm{ok} \\ $$

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