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

2(cos(45))^4 = (1/2) ร— (๐›•/(x ร— 2)) + 1 โˆ’ (2/2) How much the x is?

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

Answered by MJS_new last updated on 09/Feb/22

2(cos(45))^4 =(1/2)ร—(ฯ„/(xร—2))+1โˆ’(2/2) 2(cos(45))^4 โˆ’1+(2/2) =(1/2)ร—(ฯ„/(xร—2)) (2(cos(45))^4 โˆ’1+(2/2) )รท(1/2)=(ฯ„/(xร—2)) (((2(cos(45))^4 โˆ’1+(2/2) )รท(1/2))/ฯ„)=(1/(xร—2)) xร—2ร—(((2(cos(45))^4 โˆ’1+(2/2) )รท(1/2))/ฯ„)=1 x=(1/(2ร—(((2(cos(45))^4 โˆ’1+(2/2) )รท(1/2))/ฯ„)))

$$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} \:=\frac{\mathrm{1}}{\mathrm{2}}ร—\frac{\tau}{{x}ร—\mathrm{2}}+\mathrm{1}โˆ’\frac{\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} โˆ’\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}ร—\frac{\tau}{{x}ร—\mathrm{2}} \\ $$$$\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} โˆ’\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\tau}{{x}ร—\mathrm{2}} \\ $$$$\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} โˆ’\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}=\frac{\mathrm{1}}{{x}ร—\mathrm{2}} \\ $$$${x}ร—\mathrm{2}ร—\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} โˆ’\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}ร—\frac{\left(\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} โˆ’\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}\:\right)\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{2}}}{\tau}} \\ $$

Answered by Bagus1003 last updated on 09/Feb/22

cos(45)=((โˆš2)/2) (((โˆš2)/2))^2 =(2/4)=(1/2) ((1/2))^2 =(1/4) (((โˆš2)/2))^4 =(1/4) 2ร—(1/4)=(1/2) (1/2)=(1/2)ร—(ฯ„/(xร—2))+1โˆ’(2/2) 0.5=0.5ร—(ฯ„/(xร—2)) 0.5/0.5=1 Substitute them 1=(ฯ„/(xร—2)) ฯ„=2ฯ€ 1=((2ฯ€)/(xร—2)) (xร—2)1=((2ฯ€)/(xร—2))ร—(xร—2) xร—2=2ร—ฯ€ 2/xร—2=2ร—ฯ€/2 2/2=1 The result of x is x=3.14..=ฯ€

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

Commented by MJS_new last updated on 09/Feb/22

LOL! Thank you for making it clear!

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

Commented by Bagus1003 last updated on 09/Feb/22

No problem dude

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

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