(cos 4x+1)(cos 2x+1)(cos x+1)= (1/8) 0 ≤ x ≤ 2π


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Question Number 89079 by john santu last updated on 15/Apr/20

$(\cos 4 x + 1) (\cos 2 x + 1) (\cos x + 1) = \frac{1}{8}$ $0 ⩽ x ⩽ 2 π$
	Answered by TANMAY PANACEA. last updated on 15/Apr/20

	$2 {c o s}^{2} 2 x . 2 {c o s}^{2} x . 2 {c o s}^{2} \frac{x}{2} = \frac{1}{8}$ $8 p^{2} = \frac{1}{8}$ $p = c o s \frac{x}{2} . c o s x . c o s 2 x$ $2 s i n \frac{x}{2} . p = s i n x c o s x c o s 2 x$ $2^{2} s i n \frac{x}{2} . p = s i n 2 x . c o s 2 x$ $2^{3} s i n \frac{x}{2} . p = s i n 4 x$ $p = \frac{s i n 4 x}{8 s i n \frac{x}{2}}$ $s o 8 p^{2} = \frac{1}{8}$ $8 {(\frac{s i n 4 x}{8 s i n \frac{x}{2}})}^{2} = \frac{1}{8}$ ${s i n}^{2} 4 x = {s i n}^{2} \frac{x}{2}$ $2 {s i n}^{2} 4 x = 2 {s i n}^{2} \frac{x}{2}$ $1 - 2 {s i n}^{2} 4 x = 1 - 2 {s i n}^{2} \frac{x}{2}$ $c o s 8 x = c o s x$ $8 x = 2 n π \pm x$ $7 x = 2 n π o r 9 x = 2 n π$ $x = \frac{2 π}{7} x = \frac{2 π}{9}$ $x = \frac{4 π}{7} x = \frac{4 π}{9}$ $x = \frac{6 π}{7} x = \frac{6 π}{9}$ $x = \frac{8 π}{7} x = \frac{8 π}{9}$ $x = \frac{10 π}{7} x = \frac{10 π}{9}$ $x = \frac{14 π}{7} x = \frac{14 π}{9}$ $- x = \frac{16 π}{9}$ $x = - x = \frac{18 π}{9}$ $p l s c h e c k$
	Commented by john santu last updated on 15/Apr/20

	$y e s s i r . o u r c o r r e c t i s s a m e$
	Answered by john santu last updated on 15/Apr/20

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