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Question Number 131421 by Lordose last updated on 04/Feb/21
Solve using trig method 8x^3 −4x^2 −4x+1=0
$$\mathrm{Solve}\:\mathrm{using}\:\mathrm{trig}\:\mathrm{method} \\ $$$$\:\:\mathrm{8x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}=\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 04/Feb/21
x=y+(1/6) 8y^3 +(1/(27))+4y^2 +((2y)/3)−4y^2 −((4y)/3)−(1/9)−4y−(2/3)+1=0 ⇒8y^3 −((14y)/3)+(7/(27))=0 y=Rcosθ ⇒ cos^3 θ−(7/(12R^2 ))cosθ+(7/(216R^3 ))=0 ((27)/( 64(√7))) Comparing with cos^3 θ−(3/4)cosθ−(1/4)cos3θ=0 (7/(12R^2 ))=(3/4)⇒R=±((√7)/3) cos3θ=±((27)/(64(√7))) θ=kπ+(1/3)cos^(−1) ±(((27)/(64(√7)))) ⇒Rcosθ=((±(√7))/3)cos(kπ+(1/3)cos^(−1) (±((27)/(64(√7))))) x=±((√7)/3)cos(kπ+(1/3)cos^(−1) (±((27)/(64(√7)))))−(1/6)
$${x}={y}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{8}{y}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{27}}+\mathrm{4}{y}^{\mathrm{2}} +\frac{\mathrm{2}{y}}{\mathrm{3}}−\mathrm{4}{y}^{\mathrm{2}} −\frac{\mathrm{4}{y}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{4}{y}−\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{8}{y}^{\mathrm{3}} −\frac{\mathrm{14}{y}}{\mathrm{3}}+\frac{\mathrm{7}}{\mathrm{27}}=\mathrm{0} \\ $$$${y}={Rcos}\theta\:\:\:\:\Rightarrow\:\:\:{cos}^{\mathrm{3}} \theta−\frac{\mathrm{7}}{\mathrm{12}{R}^{\mathrm{2}} }{cos}\theta+\frac{\mathrm{7}}{\mathrm{216}{R}^{\mathrm{3}} }=\mathrm{0}\:\:\:\:\:\:\:\frac{\mathrm{27}}{\:\mathrm{64}\sqrt{\mathrm{7}}} \\ $$$${Comparing}\:{with}\:{cos}^{\mathrm{3}} \theta−\frac{\mathrm{3}}{\mathrm{4}}{cos}\theta−\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}\theta=\mathrm{0} \\ $$$$\frac{\mathrm{7}}{\mathrm{12}{R}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{R}=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:{cos}\mathrm{3}\theta=\pm\frac{\mathrm{27}}{\mathrm{64}\sqrt{\mathrm{7}}} \\ $$$$\theta={k}\pi+\frac{\mathrm{1}}{\mathrm{3}}{cos}^{−\mathrm{1}} \pm\left(\frac{\mathrm{27}}{\mathrm{64}\sqrt{\mathrm{7}}}\right)\:\Rightarrow{Rcos}\theta=\frac{\pm\sqrt{\mathrm{7}}}{\mathrm{3}}{cos}\left({k}\pi+\frac{\mathrm{1}}{\mathrm{3}}{cos}^{−\mathrm{1}} \left(\pm\frac{\mathrm{27}}{\mathrm{64}\sqrt{\mathrm{7}}}\right)\right) \\ $$$${x}=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}{cos}\left({k}\pi+\frac{\mathrm{1}}{\mathrm{3}}{cos}^{−\mathrm{1}} \left(\pm\frac{\mathrm{27}}{\mathrm{64}\sqrt{\mathrm{7}}}\right)\right)−\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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