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Question Number 65723 by bshahid010@gmail.com last updated on 02/Aug/19

	Answered by Tanmay chaudhury last updated on 03/Aug/19
	$7a=2π 4a=2π−3a cos(4a)=cos(2π−3a) (2cos^2 2a−1)=cos3π cosa=c so cos2a=2c^2 −1 cos3a=4c^3 −3c {2(2c^2 −1)^2 −1}=(4c^3 −3c) {2(4c^4 −4c^2 +1)−1}=(4c^3 −3c) 8c^4 −8c^2 +1=4c^3 −3c 8c^4 −4c^3 −8c^2 +3c+1=0 8c^4 −8c^3 +4c^3 −4c^2 −4c^2 +4c−c+1=0 8c^3 (c−1)+4c^2 (c−1)−4c(c−1)−1(c−1)=0 (c−1)(8c^3 +4c^2 −4c−1)=0 8c^3 +4c^2 −4c−1=0 c=cosa=cos(((2π)/7)) [since 7a=2π] cos(((2π)/7)) is a root of 8x^3 +4x^2 −4x−1=0$
	$7 a = 2 π$ $4 a = 2 π - 3 a$ $c o s (4 a) = c o s (2 π - 3 a)$ $(2 {c o s}^{2} 2 a - 1) = c o s 3 π$ $c o s a = c s o c o s 2 a = 2 c^{2} - 1$ $c o s 3 a = 4 c^{3} - 3 c$ ${2 {(2 c^{2} - 1)}^{2} - 1} = (4 c^{3} - 3 c)$ ${2 (4 c^{4} - 4 c^{2} + 1) - 1} = (4 c^{3} - 3 c)$ $8 c^{4} - 8 c^{2} + 1 = 4 c^{3} - 3 c$ $8 c^{4} - 4 c^{3} - 8 c^{2} + 3 c + 1 = 0$ $8 c^{4} - 8 c^{3} + 4 c^{3} - 4 c^{2} - 4 c^{2} + 4 c - c + 1 = 0$ $8 c^{3} (c - 1) + 4 c^{2} (c - 1) - 4 c (c - 1) - 1 (c - 1) = 0$ $(c - 1) (8 c^{3} + 4 c^{2} - 4 c - 1) = 0$ $8 c^{3} + 4 c^{2} - 4 c - 1 = 0$ $c = c o s a = c o s (\frac{2 π}{7}) [s i n c e 7 a = 2 π]$ $c o s (\frac{2 π}{7}) i s a r o o t o f$ $8 x^{3} + 4 x^{2} - 4 x - 1 = 0$
	Commented by bshahid010@gmail.com last updated on 03/Aug/19

	$thank you$
	Commented by Tanmay chaudhury last updated on 03/Aug/19

	$m o s t w e l c o m e$
	Commented by MJS last updated on 03/Aug/19

	$but \cos \frac{π}{7} d o e s solve the equation . . .$ $the other roots are \cos \frac{3 π}{7} and \cos \frac{5 π}{7}$ $8 (x - \cos \frac{π}{7}) (x - \cos \frac{3 π}{7}) (x - \cos \frac{5 π}{7}) \approx$ $\approx 8 (x - . 900 968 868) (x - . 222 520 934) (x + . 623 489 801) \approx$ $\approx 8 x^{3} - 4 x^{2} - 4 x + 1$
	Commented by MJS last updated on 05/Aug/19

	$I still {don}^{'} t understand why you did what you$ $did . it works exactly the sane way with the$ $given value$ $7 a = π$ $4 a = π - 3 a$ $\cos 4 a = - \cos 3 a$ $a = \arccos c$ $8 c^{4} - 8 c^{2} + 1 = 3 c - 4 c^{3}$ $8 c^{4} + 4 c^{3} - 8 c^{2} - 3 c + 1 = 0$ $(c + 1) (8 c^{3} - 4 c^{2} - 4 c + 1) = 0$ $a = \frac{π}{7} \Rightarrow \cos \frac{π}{7} is a root of the given equation$
	Commented by Tanmay chaudhury last updated on 05/Aug/19

	$s i r y o u r a n s w e r a l s o c o r r e c t . . . y o u h a v e t a k e n$ $7 a = π a n d i h a v e t a k e n 7 a = 2 π$ $b o t h a r e c o r r e c t . . . e q u a t i o n a r e d i f f e r e n t . . .$ $i t h o u g h t 7 a = 2 π s o i d i d i t . . .$ $y o u r a n z w e r i s a s p e r g i v v e n c o n d i t i o n i n a u e s t i o n$ $g$
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