2-cos-45-4-1-2-x-2-1-2-2-How-much-the-x-is-

Question Number 165851 by Bagus1003 last updated on 09/Feb/22

$$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\boldsymbol{\tau}}{{x}\:×\:\mathrm{2}}\:+\:\mathrm{1}\:−\:\frac{\mathrm{2}}{\mathrm{2}} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$

Answered by MJS_new last updated on 09/Feb/22

$$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} \:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\tau}{{x}×\mathrm{2}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\tau}{{x}×\mathrm{2}} \\ $$$$\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\tau}{{x}×\mathrm{2}} \\ $$$$\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}=\frac{\mathrm{1}}{{x}×\mathrm{2}} \\ $$$${x}×\mathrm{2}×\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}×\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}} \\ $$

Answered by Bagus1003 last updated on 09/Feb/22

$$\mathrm{cos}\left(\mathrm{45}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\tau}{{x}×\mathrm{2}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{0}.\mathrm{5}=\mathrm{0}.\mathrm{5}×\frac{\tau}{{x}×\mathrm{2}} \\ $$$$\mathrm{0}.\mathrm{5}/\mathrm{0}.\mathrm{5}=\mathrm{1} \\ $$$${Substitute}\:{them} \\ $$$$\mathrm{1}=\frac{\tau}{{x}×\mathrm{2}} \\ $$$$\tau=\mathrm{2}\pi \\ $$$$\mathrm{1}=\frac{\mathrm{2}\pi}{{x}×\mathrm{2}} \\ $$$$\left({x}×\mathrm{2}\right)\mathrm{1}=\frac{\mathrm{2}\pi}{{x}×\mathrm{2}}×\left({x}×\mathrm{2}\right) \\ $$$${x}×\mathrm{2}=\mathrm{2}×\pi \\ $$$$\mathrm{2}/{x}×\mathrm{2}=\mathrm{2}×\pi/\mathrm{2} \\ $$$$\mathrm{2}/\mathrm{2}=\mathrm{1} \\ $$$${The}\:{result}\:{of}\:{x}\:{is} \\ $$$${x}=\mathrm{3}.\mathrm{14}..=\pi \\ $$

Commented by MJS_new last updated on 09/Feb/22

$$\mathrm{LOL}!\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{making}\:\mathrm{it}\:\mathrm{clear}! \\ $$

Commented by Bagus1003 last updated on 09/Feb/22

$${No}\:{problem}\:{dude} \\ $$

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