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Question Number 42174 by lucha116 last updated on 19/Aug/18

cos^2 3xcos2x−cos^2 x=0

$${cos}^{\mathrm{2}} \mathrm{3}{xcos}\mathrm{2}{x}−{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18

(((1+cos6x)/2))cos2x−(((1+cos2x)/2))=0 cos2x+cos2x.cos6x−1−cos2x=0 cos2x.cos6x=1 2cos2x.cos6x=2 cos8x+cos4x=2 cos4x=k 2k^2 −1+k−2=0 2k^2 +k−3=0 2k^2 +3k−2k−3=0 k(2k+3)−1(2k+3)=0 (2k+3)(k−1)=0 k−1=0 k=1 cos4x=1=coso 4x=2nΠ x=((nΠ)/2)

$$\left(\frac{\mathrm{1}+{cos}\mathrm{6}{x}}{\mathrm{2}}\right){cos}\mathrm{2}{x}−\left(\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${cos}\mathrm{2}{x}+{cos}\mathrm{2}{x}.{cos}\mathrm{6}{x}−\mathrm{1}−{cos}\mathrm{2}{x}=\mathrm{0} \\ $$$${cos}\mathrm{2}{x}.{cos}\mathrm{6}{x}=\mathrm{1} \\ $$$$\mathrm{2}{cos}\mathrm{2}{x}.{cos}\mathrm{6}{x}=\mathrm{2} \\ $$$${cos}\mathrm{8}{x}+{cos}\mathrm{4}{x}=\mathrm{2} \\ $$$${cos}\mathrm{4}{x}={k} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}+{k}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} +{k}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}−\mathrm{2}{k}−\mathrm{3}=\mathrm{0} \\ $$$${k}\left(\mathrm{2}{k}+\mathrm{3}\right)−\mathrm{1}\left(\mathrm{2}{k}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{k}+\mathrm{3}\right)\left({k}−\mathrm{1}\right)=\mathrm{0} \\ $$$${k}−\mathrm{1}=\mathrm{0} \\ $$$${k}=\mathrm{1} \\ $$$${cos}\mathrm{4}{x}=\mathrm{1}={coso} \\ $$$$\mathrm{4}{x}=\mathrm{2}{n}\Pi \\ $$$${x}=\frac{{n}\Pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by lucha116 last updated on 20/Aug/18

thks

$${thks} \\ $$

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