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

$$1+3\sqrt{2}x-18x^2=6x\sqrt{1-9x^2}$$$$solveinR$$

Answered by MJS_new last updated on 20/Feb/22

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

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

$$let3x=\sin \theta$$$$1+\sqrt{2}\sin \theta -2{\sin }^2\theta =2\sin \theta \cos \theta$$$$\cos 2\theta +\sqrt{2}\sin \theta =\sin 2\theta$$$$\sin \theta =\frac{1}{\sqrt{2}}(\sin 2\theta -\cos 2\theta )$$$$\sin \theta =\sin (2\theta -\frac{\pi }{4})$$$$2\theta -\frac{\pi }{4}=2k\pi +\theta \Rightarrow \theta =2k\pi +\frac{\pi }{4}$$$$2\theta -\frac{\pi }{4}=(2k+1)\pi -\theta \Rightarrow \theta =\frac{2k\pi }{3}+\frac{5\pi }{12}$$$$\Rightarrow x_1=\frac{1}{3}\sin \frac{\pi }{4}=\frac{\sqrt{2}}{6}$$$$\Rightarrow x_2=\frac{1}{3}\sin \frac{5\pi }{12}=\frac{\sqrt{6}+\sqrt{2}}{12}$$$$\Rightarrow x_3=-\frac{1}{3}\sin \left(\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{6}$$$$\Rightarrow x=\frac{1}{3}\sin \left(\frac{13\pi }{12}\right)(rejected,since\cos \frac{13\pi }{12}<0)$$

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

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

Commented by MaxiMaths last updated on 20/Feb/22

$$\mathrm{ok}$$

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