Question Number 40906 by Penguin last updated on 29/Jul/18 $${f}\left({x}\right)\:=\:\mathrm{5}.\mathrm{687cosh}\left(\frac{{x}}{\mathrm{5}.\mathrm{687}}\right)−\mathrm{5}.\mathrm{687} \\ $$$${L}=\int_{-\mathrm{11}} ^{\mathrm{11}} \sqrt{\mathrm{1}+\left[{f}\:'\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 40875 by Tawa1 last updated on 28/Jul/18 $$\mathrm{2a}\:\mathrm{sin}\left(\frac{\mathrm{25}}{\mathrm{a}}\right)\:−\:\mathrm{51}\:=\:\mathrm{0},\:\:\mathrm{find}\:\mathrm{a} \\ $$ Answered by MrW3 last updated on 29/Jul/18 $${let}'{s}\:{have}\:{a}\:{look}\:{at}\:{the}\:{function}\:{f}\left({x}\right)=\frac{\mathrm{sin}\:\left({x}\right)}{{x}}. \\ $$$${we}\:{know}\:{following}\:{about}\:{it}: \\ $$$${f}\left(−{x}\right)={f}\left({x}\right)\:\Rightarrow{symmetric} \\…
Question Number 106387 by bemath last updated on 05/Aug/20 $$\mathrm{cot}\:\theta\:+\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\:=\:\mathrm{2}\: \\ $$$$\theta\:=\:? \\ $$ Answered by 1549442205PVT last updated on 05/Aug/20 $$\mathrm{cot}\:\theta\:+\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\:=\:\mathrm{2}\: \\ $$$$\Leftrightarrow\frac{\mathrm{sin}\left(\mathrm{2}\theta+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\theta\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\theta\right)}=\mathrm{2}\Leftrightarrow\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}\theta\right)=\mathrm{2sin}\theta\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\theta\right) \\…
Question Number 171870 by Mikenice last updated on 21/Jun/22 $${if}\:{x}\:{is}\:{a}\:{real}\:{number},{then} \\ $$$$\sqrt{{log}_{{e}} \:\frac{\mathrm{4}{x}−{x}^{\mathrm{2}} }{\mathrm{3}}\:}\:\: \\ $$ Answered by floor(10²Eta[1]) last updated on 21/Jun/22 $$??\:\mathrm{whats}\:\mathrm{your}\:\mathrm{question} \\…
Question Number 171858 by Mikenice last updated on 21/Jun/22 $${solve}: \\ $$$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y}\:{between}\:\mathrm{0}^{°\:} {and}\mathrm{180}^{°} \:{which}\:{satisfy}\:{the}\:{simultaneous}\:{eqn} \\ $$$${sin}\left({x}+{y}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${cos}\mathrm{2}{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 106295 by bobhans last updated on 04/Aug/20 $$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{x}}\:=\:\mathrm{4}\: \\ $$ Answered by bemath last updated on 04/Aug/20 $$\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{4sin}\:\mathrm{xcos}\:\mathrm{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}=\mathrm{sin}\:\mathrm{2x} \\ $$$$\mathrm{sin}\:\mathrm{30}°\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{30}°\mathrm{cos}\:\mathrm{x}=\mathrm{sin}\:\mathrm{2x} \\…
Question Number 106286 by bemath last updated on 04/Aug/20 $$\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)=\mathrm{arc}\:\mathrm{tan}\:\left(−\mathrm{7}\right) \\ $$$$\mathrm{for}\:\mathrm{x}\:\mathrm{real}\:\mathrm{number} \\ $$ Answered by bobhans last updated on 04/Aug/20 $$\rightarrow\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)=\mathrm{arc}\:\mathrm{tan}\:\left(−\mathrm{7}\right) \\ $$$$\mathrm{tan}\:\left(\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right)\right)=\mathrm{tan}\:\left(\mathrm{arc}\:\mathrm{tan}\:\left(−\mathrm{7}\right)\right) \\…
Question Number 106220 by bemath last updated on 03/Aug/20 $$\mathrm{find}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{45}°\right)=\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{60}°\right) \\ $$ Answered by Dwaipayan Shikari last updated on 03/Aug/20 $${sin}\left(\frac{\pi}{\mathrm{2}}−{x}+\frac{\pi}{\mathrm{4}}\right)={sin}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{4}}−{x}=\mathrm{2}{k}\pi\pm\mathrm{2}{x}+\frac{\pi}{\mathrm{3}}\:\:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$${first}\:{case}…
Question Number 106214 by john santu last updated on 03/Aug/20 $$\mathrm{solve}\::\:\mathrm{arc}\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{1}\right)=\:\mathrm{2arc}\:\mathrm{cos}\:\left(\mathrm{x}\right) \\ $$$$\mathrm{where}\:\mathrm{x}\:\mathrm{is}\:\mathrm{real}. \\ $$ Answered by bobhans last updated on 03/Aug/20 $$\Rightarrow\mathrm{cos}\:\left(\mathrm{arc}\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{1}\right)\right)=\:\mathrm{cos}\:\left(\mathrm{2}\:\mathrm{arc}\:\mathrm{cos}\:\left(\mathrm{x}\right)\right) \\ $$$$\Rightarrow\:\mathrm{x}−\mathrm{1}\:=\:\mathrm{2}\:\mathrm{cos}\:^{\mathrm{2}}…
Question Number 106196 by john santu last updated on 03/Aug/20 Commented by john santu last updated on 03/Aug/20 Commented by john santu last updated on…