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cos-4x-1-cos-2x-1-cos-x-1-1-8-0-x-2pi-

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Question Number 89079 by john santu last updated on 15/Apr/20
(cos 4x+1)(cos 2x+1)(cos x+1)= (1/8) 0 ≤ x ≤ 2π
$$\left(\mathrm{cos}\:\mathrm{4}{x}+\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{cos}\:{x}+\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{2}\pi \\ $$
Answered by TANMAY PANACEA. last updated on 15/Apr/20
2cos^2 2x.2cos^2 x.2cos^2 (x/2)=(1/8) 8p^2 =(1/8) p=cos(x/2).cosx.cos2x 2sin(x/2).p=sinxcosxcos2x 2^2 sin(x/2).p=sin2x.cos2x 2^3 sin(x/2).p=sin4x p=((sin4x)/(8sin(x/2))) so 8p^2 =(1/8) 8(((sin4x)/(8sin(x/2))))^2 =(1/8) sin^2 4x=sin^2 (x/2) 2sin^2 4x=2sin^2 (x/2) 1−2sin^2 4x=1−2sin^2 (x/2) cos8x=cosx 8x=2nπ±x 7x=2nπ or 9x=2nπ x=((2π)/7) x=((2π)/9) x=((4π)/7) x=((4π)/9) x=((6π)/7) x=((6π)/9) x=((8π)/7) x=((8π)/9) x=((10π)/7) x=((10π)/9) x=((14π)/7) x=((14π)/9) − x=((16π)/9) x=− x=((18π)/9) pls check
$$\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{x}.\mathrm{2}{cos}^{\mathrm{2}} {x}.\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{8}{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${p}={cos}\frac{{x}}{\mathrm{2}}.{cosx}.{cos}\mathrm{2}{x} \\ $$$$\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}.{p}={sinxcosxcos}\mathrm{2}{x} \\ $$$$\mathrm{2}^{\mathrm{2}} {sin}\frac{{x}}{\mathrm{2}}.{p}={sin}\mathrm{2}{x}.{cos}\mathrm{2}{x} \\ $$$$\mathrm{2}^{\mathrm{3}} {sin}\frac{{x}}{\mathrm{2}}.{p}={sin}\mathrm{4}{x} \\ $$$${p}=\frac{{sin}\mathrm{4}{x}}{\mathrm{8}{sin}\frac{{x}}{\mathrm{2}}} \\ $$$${so}\:\mathrm{8}{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{8}\left(\frac{{sin}\mathrm{4}{x}}{\mathrm{8}{sin}\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${sin}^{\mathrm{2}} \mathrm{4}{x}={sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}{x}=\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}{x}=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$${cos}\mathrm{8}{x}={cosx} \\ $$$$\mathrm{8}{x}=\mathrm{2}{n}\pi\pm{x} \\ $$$$\mathrm{7}{x}=\mathrm{2}{n}\pi\:\:\:\:{or}\:\:\:\:\mathrm{9}{x}=\mathrm{2}{n}\pi \\ $$$${x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\:\:\:\:\:{x}=\frac{\mathrm{2}\pi}{\mathrm{9}} \\ $$$${x}=\frac{\mathrm{4}\pi}{\mathrm{7}}\:\:\:\:\:\:{x}=\frac{\mathrm{4}\pi}{\mathrm{9}} \\ $$$${x}=\frac{\mathrm{6}\pi}{\mathrm{7}}\:\:\:\:{x}=\frac{\mathrm{6}\pi}{\mathrm{9}} \\ $$$${x}=\frac{\mathrm{8}\pi}{\mathrm{7}}\:\:\:\:\:{x}=\frac{\mathrm{8}\pi}{\mathrm{9}} \\ $$$${x}=\frac{\mathrm{10}\pi}{\mathrm{7}}\:\:\:\:{x}=\frac{\mathrm{10}\pi}{\mathrm{9}} \\ $$$${x}=\frac{\mathrm{14}\pi}{\mathrm{7}}\:\:\:\:\:\:{x}=\frac{\mathrm{14}\pi}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:{x}=\frac{\mathrm{16}\pi}{\mathrm{9}} \\ $$$${x}=−\:\:\:\:\:\:\:{x}=\frac{\mathrm{18}\pi}{\mathrm{9}} \\ $$$${pls}\:{check} \\ $$
Commented by john santu last updated on 15/Apr/20
yes sir. our correct is same
$${yes}\:{sir}.\:{our}\:{correct}\:{is}\:{same} \\ $$
Answered by john santu last updated on 15/Apr/20

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