Let S ⊂ (0, π) denote the set of values of x satisfying the equation 8^(1+∣cos x∣+cos^2 x+∣cos^3 x∣+... to ∞) = 4^3 then S =


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Question Number 27203 by julli deswal last updated on 03/Jan/18

$Let S \subset (0, π) denote the set of values of x$ $satisfying the equation$ $8^{1 + ∣ \cos x ∣ + \cos^{2} x + ∣ \cos^{3} x ∣ + . . . to \infty} = 4^{3}$ $then S =$
	Answered by Tinkutara last updated on 03/Jan/18
	$8^(1/(1−∣cos x∣)) =8^2 1−∣cos x∣=(1/2) cos x=±(1/2) S={(π/3),((2π)/3)}$
	$8^{\frac{1}{1 - ∣ \cos x ∣}} = 8^{2}$ $1 - ∣ \cos x ∣= \frac{1}{2}$ $\cos x = \pm \frac{1}{2}$ $S = {\frac{π}{3}, \frac{2 π}{3}}$
	Answered by mrW1 last updated on 03/Jan/18
	$4^3 =8^2 let t=∣cos x∣ 1+t+t^2 +t^3 +....=(1/(1−t))=2 ⇒t=(1/2) ⇒cos x=±(1/2) ⇒x=(π/3) or ((2π)/3) ⇒S={(π/3) ,((2π)/3)}$
	$4^{3} = 8^{2}$ $l e t t =∣ \cos x ∣$ $1 + t + t^{2} + t^{3} + . . . . = \frac{1}{1 - t} = 2$ $\Rightarrow t = \frac{1}{2}$ $\Rightarrow \cos x = \pm \frac{1}{2}$ $\Rightarrow x = \frac{π}{3} o r \frac{2 π}{3}$ $\Rightarrow S = {\frac{π}{3}, \frac{2 π}{3}}$
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