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Question Number 136494 by liberty last updated on 22/Mar/21

$$5(\sqrt{1-x}+\sqrt{1+x})=6x+8\sqrt{1-x^2}$$

Answered by MJS_new last updated on 22/Mar/21

$$\mathrm{I}\mathrm{get}2\mathrm{solutions}$$$$x_1=\frac{24}{25}$$$$x_2=\cos \frac{\pi +2\arcsin \frac{\sqrt{2}}{10}}{3}\approx .415962301$$$$\mathrm{the}\mathrm{path}\mathrm{is}:\mathrm{let}t=\sqrt{1-x}\mathrm{then}\mathrm{transform},$$$$\mathrm{square}\mathrm{and}\mathrm{solve}\mathrm{the}{4}^{\mathrm{th}}\mathrm{degree}\mathrm{for}t,\mathrm{then}$$$$\mathrm{test}\mathrm{the}4\mathrm{solutions}$$$$\mathrm{you}\mathrm{can}\mathrm{also}\mathrm{try}\mathrm{with}x=\cos 2t\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{much}$$$$\mathrm{easier}...$$

Answered by mr W last updated on 22/Mar/21

$$5(\sqrt{1-x}+\sqrt{1+x})=6(x+1)-6+8\sqrt{1-x^2}$$$$letp=\sqrt{1+x}⩾0$$$$q=\sqrt{1-x}⩾0$$$$\Rightarrow p^2=1+x$$$$\Rightarrow q^2=1-x$$$$\Rightarrow p^2+q^2=2$$$$\Rightarrow p=\sqrt{2}\cos \theta ⩾0$$$$\Rightarrow q=\sqrt{2}\sin \theta ⩾0$$$$$$$$5(p+q)=6p^2-6+8pq$$$$5\sqrt{2}(\cos \theta +\sin \theta )=2(3\cos 2\theta +4\sin 2\theta )$$$$\frac{\sqrt{2}}{2}(\cos \theta +\sin \theta )=\frac{3}{5}\cos 2\theta +\frac{4}{5}\sin 2\theta$$$$let\alpha ={\tan }^{-1}\frac{4}{3}$$$$\Rightarrow \cos (\theta -\frac{\pi }{4})=\cos (2\theta -\alpha )$$$$\Rightarrow 2\theta -\alpha =2k\pi \pm (\theta -\frac{\pi }{4})$$$$\Rightarrow 2\theta -\alpha =2k\pi +\theta -\frac{\pi }{4}$$$$\Rightarrow {\theta }_1=2k\pi -\frac{\pi }{4}+\alpha withk=0$$$$or$$$$\Rightarrow 2\theta -\alpha =2k\pi -\theta +\frac{\pi }{4}$$$$\Rightarrow 3\theta =2k\pi +\frac{\pi }{4}+\alpha$$$$\Rightarrow {\theta }_{2,3,4}=\frac{2k\pi }{3}+\frac{\pi }{12}+\frac{\alpha }{3}withk=0,1,2$$$$since\cos \theta ⩾0,\sin \theta ⩾0,onlyvalid:$$$${\theta }_1=-\frac{\pi }{4}+\alpha$$$${\theta }_2=\frac{\pi }{12}+\frac{\alpha }{3}$$$$x=p^2-1$$$$={\left(\sqrt{2}\cos \theta \right)}^2-1$$$$=2{\cos }^2(-\frac{\pi }{4}+{\tan }^{-1}\frac{4}{3})-1=\frac{24}{25}=0.96$$$$or$$$$=2{\cos }^2(\frac{\pi }{12}+\frac{1}{3}{\tan }^{-1}\frac{4}{3})-1\approx 0.415962$$

Answered by ajfour last updated on 23/Mar/21

$$x=\cos 2\theta$$$$5\sqrt{2}(\sin \theta +\cos \theta )=6\cos 2\theta +8\sin 2\theta$$$$10\sin (\theta +\frac{\pi }{4})=10\sin (2\theta +{\tan }^{-1}\frac{3}{4})$$$$firstletthetwoarguments$$$$besame\Rightarrow$$$$\theta =\frac{\pi }{4}-{\tan }^{-1}\frac{3}{4}$$$$x=\cos (\frac{\pi }{2}-{2\tan }^{-1}\frac{3}{4})$$$$=\sin \left({2\tan }^{-1}\frac{3}{4}\right)$$$$=2\sin \left({\tan }^{-1}\frac{3}{4}\right)\cos \left({\tan }^{-1}\frac{3}{4}\right)$$$$=2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25}.$$

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