Solve-using-trig-method-8x-3-4x-2-4x-1-0-

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

Solve using trig method

{8 x}^{3} - {4 x}^{2} - 4 x + 1 = 0

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

x = y + \frac{1}{6}

8 y^{3} + \frac{1}{27} + 4 y^{2} + \frac{2 y}{3} - 4 y^{2} - \frac{4 y}{3} - \frac{1}{9} - 4 y - \frac{2}{3} + 1 = 0

\Rightarrow 8 y^{3} - \frac{14 y}{3} + \frac{7}{27} = 0

y = R c o s θ \Rightarrow {c o s}^{3} θ - \frac{7}{12 R^{2}} c o s θ + \frac{7}{216 R^{3}} = 0 \frac{27}{64 \sqrt{7}}

C o m p a r i n g w i t h {c o s}^{3} θ - \frac{3}{4} c o s θ - \frac{1}{4} c o s 3 θ = 0

\frac{7}{12 R^{2}} = \frac{3}{4} \Rightarrow R = \pm \frac{\sqrt{7}}{3} c o s 3 θ = \pm \frac{27}{64 \sqrt{7}}

θ = k π + \frac{1}{3} {c o s}^{- 1} \pm (\frac{27}{64 \sqrt{7}}) \Rightarrow R c o s θ = \frac{\pm \sqrt{7}}{3} c o s (k π + \frac{1}{3} {c o s}^{- 1} (\pm \frac{27}{64 \sqrt{7}}))

x = \pm \frac{\sqrt{7}}{3} c o s (k π + \frac{1}{3} {c o s}^{- 1} (\pm \frac{27}{64 \sqrt{7}})) - \frac{1}{6}