cos^2 x+cos^2 2x=sin^2 3x Solve the equation. Please help ASAP


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Question Number 108143 by abony1303 last updated on 15/Aug/20

$\cos^{2} x + \cos^{2} 2 x = \sin^{2} 3 x$ $Solve the equation .$ $Please help ASAP$
	Answered by ajfour last updated on 15/Aug/20

	$1 + \cos 2 x + 2 \cos^{2} 2 x$ $= 2 \sin^{2} x {(3 - 4 \sin^{2} x)}^{2}$ $l e t \cos 2 x = t$ $\Rightarrow 1 + t + 2 t^{2} = (1 - t) {(3 - 2 + 2 t)}^{2}$ $\Rightarrow$ $t (2 t + 1) + t {(2 t + 1)}^{2} - {(2 t + 1)}^{2} + 1 = 0$ $\Rightarrow (2 t + 1) (2 t^{2} + t + t - 2 t - 1) + 1 = 0$ $\Rightarrow (2 t + 1) (2 t^{2} - 1) + 1 = 0$ $\Rightarrow 4 t^{3} + 2 t^{2} - 2 t = 0$ $\Rightarrow t (2 t^{2} + t - 1) = 0$ $\Rightarrow t (t + 1) ((2 t - 1) = 0$ $\Rightarrow t = \cos 2 x = 0, \frac{1}{2}, - 1 .$
	Answered by 1549442205PVT last updated on 15/Aug/20
	$cos^2 x+cos^2 2x=sin^2 3x ⇔1−cos2x+1+cos4x=1−cos6x ⇔cos6x−cos2x+cos4x+1=0 ⇔−2sin4xsin2x+2cos^2 2x=0 ⇔2sin^2 2xcos2x+cos^2 2x=0 ⇔cos2x(2sin^2 2x+cos2x)=0 ⇔cos2x(−2cos^2 2x+cos2x+2)=0 i)cos2x=0⇔2x=(π/2)+kπ⇔x=(π/4)+((kπ)/2) ii)−2cos^2 2x+cos2x+2=0 ⇔2cos^2 2x−cos2x−2=0 Δ=1+4.2.2=17 •cos2x=((1+(√(17)))/4)>1⇒is rejected •cos2x=((1−(√(17)))/4)⇒2x=cos^(−1) (((1−(√(17)))/4))+2kπ ⇔x=(1/2)cos^(−1) (((1−(√(17)))/4))+kπ Thus,the roots of given equation are x∈{(𝛑/4)+((m𝛑)/2);(1/2)cos^(−1) (((1−(√(17)))/4))+k𝛑}$
	$\cos^{2} x + \cos^{2} 2 x = \sin^{2} 3 x$ $\Leftrightarrow 1 - \cos 2 x + 1 + \cos 4 x = 1 - \cos 6 x$ $\Leftrightarrow \cos 6 x - \cos 2 x + \cos 4 x + 1 = 0$ $\Leftrightarrow - 2 \sin 4 xsin 2 x + {2 \cos}^{2} 2 x = 0$ $\Leftrightarrow {2 \sin}^{2} 2 xcos 2 x + \cos^{2} 2 x = 0$ $\Leftrightarrow \cos 2 x ({2 \sin}^{2} 2 x + \cos 2 x) = 0$ $\Leftrightarrow \cos 2 x (- {2 \cos}^{2} 2 x + \cos 2 x + 2) = 0$ $i) \cos 2 x = 0 \Leftrightarrow 2 x = \frac{π}{2} + k π \Leftrightarrow x = \frac{π}{4} + \frac{k π}{2}$ $ii) - {2 \cos}^{2} 2 x + \cos 2 x + 2 = 0$ $\Leftrightarrow {2 \cos}^{2} 2 x - \cos 2 x - 2 = 0$ $Δ = 1 + 4.2.2 = 17$ $∙ \cos 2 x = \frac{1 + \sqrt{17}}{4} > 1 \Rightarrow is rejected$ $∙ \cos 2 x = \frac{1 - \sqrt{17}}{4} \Rightarrow 2 x = \cos^{- 1} (\frac{1 - \sqrt{17}}{4}) + 2 k π$ $\Leftrightarrow x = \frac{1}{2} \cos^{- 1} (\frac{1 - \sqrt{17}}{4}) + k π$ $Thus, the roots of given equation are$ $x \in {\frac{π}{4} + \frac{m π}{2}; \frac{1}{2} \cos^{- 1} (\frac{1 - \sqrt{17}}{4}) + k π}$
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