Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com