8x^3 −6x+1=0 solve for x


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Question Number 166378 by MathsFan last updated on 19/Feb/22

$8 x^{3} - 6 x + 1 = 0$ $s o l v e f o r x$
	Answered by mr W last updated on 19/Feb/22

	$3 x - 4 x^{3} = \frac{1}{2}$ $w e k n o w 3 \sin θ - 4 \sin^{3} θ = \sin 3 θ, s o$ $i f \sin 3 θ = \frac{1}{2}, i . e . 3 θ = 2 k π + \sin^{- 1} \frac{1}{2}$ $t h e n x = \sin θ = \sin (\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} \frac{1}{2})$ $\Rightarrow x = \sin (\frac{2 k π}{3} + \frac{π}{18}) w i t h k = 0, 1, 2$ $x_{1} = \sin \frac{π}{18}$ $x_{2} = \sin (\frac{2 π}{3} + \frac{π}{18}) = \sin \frac{5 π}{18}$ $x_{3} = \sin (\frac{4 π}{3} + \frac{π}{18}) = - \sin \frac{7 π}{18}$
	Commented by MathsFan last updated on 19/Feb/22

	$t h a n k y o u s i r$ $b u t c a n {c a n a d o}^{'} s f o r m u l a b e a p p l i e d$ $t o t h i s q u e s t i o n ?$
	Commented by mr W last updated on 19/Feb/22

	$n o .$
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