x^3 −12x^2 +27x−17=0 Let x=t+4 t^3 −21t−37=0 The Trigonometric Solution gives these: x_1 =4−2(√7)cos ((π+2sin^(−1) ((37(√7))/(98)))/6) x_2 =4−2(√7)sin ((sin^(−1) ((37(√7))/(98)))/3) x_3 =4+2(√7)sin ((π+sin^(−1) ((37(√7))/(98)))/3) Prove these identities: x_1 =2−((1+2sin (π/(18)))/(2cos (π/9))) x_2 =2+((1+2cos (π/9))/(2cos ((2π)/9))) x_3 =((1+2(√3)sin ((2π)/9))/(2sin (π/(18))))


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Question Number 207434 by Frix last updated on 15/May/24

$x^{3} - 12 x^{2} + 27 x - 17 = 0$ $Let x = t + 4$ $t^{3} - 21 t - 37 = 0$ $The Trigonometric Solution gives these :$ $x_{1} = 4 - 2 \sqrt{7} \cos \frac{π + {2 \sin}^{- 1} \frac{37 \sqrt{7}}{98}}{6}$ $x_{2} = 4 - 2 \sqrt{7} \sin \frac{\sin^{- 1} \frac{37 \sqrt{7}}{98}}{3}$ $x_{3} = 4 + 2 \sqrt{7} \sin \frac{π + \sin^{- 1} \frac{37 \sqrt{7}}{98}}{3}$ $Prove these identities :$ $x_{1} = 2 - \frac{1 + 2 \sin \frac{π}{18}}{2 \cos \frac{π}{9}}$ $x_{2} = 2 + \frac{1 + 2 \cos \frac{π}{9}}{2 \cos \frac{2 π}{9}}$ $x_{3} = \frac{1 + 2 \sqrt{3} \sin \frac{2 π}{9}}{2 \sin \frac{π}{18}}$
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