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Question Number 131421 by Lordose last updated on 04/Feb/21

$$\mathrm{Solve}\mathrm{using}\mathrm{trig}\mathrm{method}$$$${8\mathrm{x}}^3-{4\mathrm{x}}^2-4\mathrm{x}+1=0$$

Commented by Dwaipayan Shikari last updated on 04/Feb/21

$$x=y+\frac{1}{6}$$$$8y^3+\frac{1}{27}+4y^2+\frac{2y}{3}-4y^2-\frac{4y}{3}-\frac{1}{9}-4y-\frac{2}{3}+1=0$$$$\Rightarrow 8y^3-\frac{14y}{3}+\frac{7}{27}=0$$$$y=Rcos\theta \Rightarrow {cos}^3\theta -\frac{7}{12R^2}cos\theta +\frac{7}{216R^3}=0\frac{27}{64\sqrt{7}}$$$$Comparingwith{cos}^3\theta -\frac{3}{4}cos\theta -\frac{1}{4}cos3\theta =0$$$$\frac{7}{12R^2}=\frac{3}{4}\Rightarrow R=\pm \frac{\sqrt{7}}{3}cos3\theta =\pm \frac{27}{64\sqrt{7}}$$$$\theta =k\pi +\frac{1}{3}{cos}^{-1}\pm \left(\frac{27}{64\sqrt{7}}\right)\Rightarrow Rcos\theta =\frac{\pm \sqrt{7}}{3}cos(k\pi +\frac{1}{3}{cos}^{-1}(\pm \frac{27}{64\sqrt{7}}))$$$$x=\pm \frac{\sqrt{7}}{3}cos(k\pi +\frac{1}{3}{cos}^{-1}(\pm \frac{27}{64\sqrt{7}}))-\frac{1}{6}$$

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