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Question Number 188226 by Shrinava last updated on 26/Feb/23

$$\mathrm{If}\mathrm{\Omega }=\sum _{\mathbf{cyc}}\frac{\sin (\mathrm{A}-\frac{\pi }{6})}{\cos (\mathrm{B}-\frac{\pi }{6})\cos (\mathrm{C}-\frac{\pi }{6})}\mathrm{in}△\mathrm{ABC}$$$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$${\mathrm{x}}^4-4\mathrm{\Omega }{\mathrm{x}}^3+6\mathrm{\Omega }{\mathrm{x}}^2-4\mathrm{\Omega }\mathrm{x}+1=0$$

Answered by aleks041103 last updated on 27/Feb/23

$$supposeABCequilateral$$$$A=B=C=\frac{\pi }{3}$$$$\Rightarrow \mathrm{\Omega }=\frac{3sin\left(\pi /6\right)}{{cos}^2\left(\pi /6\right)}=3\frac{\frac{1}{2}}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{3}{2}\frac{4}{3}=2$$$$\Rightarrow x^4-8x^3+12x^2-8x+1=0$$$$x=0isnotsol$$$$\Rightarrow x^2+\frac{1}{x^2}-8(x+\frac{1}{x})+12=0$$$$y=x+\frac{1}{x}\Rightarrow y^2=x^2+\frac{1}{x^2}+2$$$$\Rightarrow y^2-2-8y+12=0$$$$y^2-8y+10=0$$$$y_{1,2}=\frac{8\pm \sqrt{64-4.10}}{2}=\frac{8\pm 2\sqrt{6}}{2}=4\pm \sqrt{6}\approx 1.55,6.45$$$$x^2-yx+1=0$$$$\Rightarrow x_{1,2,3,4}=\frac{y_{1,2}\pm \sqrt{y_{1,2}^2-4}}{2}=\frac{y_{1,2}\pm \sqrt{6-8y_{1,2}}}{2}$$$$but6-8y_{1,2}<0\Rightarrow x\notin \mathbb{R}$$

Commented by Shrinava last updated on 28/Feb/23

$$\mathrm{perfect}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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