Question Number 147992 by Sozan last updated on 24/Jul/21 Commented by Sozan last updated on 25/Jul/21 $${who}\:{is}\:{can}\:{solve}\:{this}\:{please}\:? \\ $$ Answered by mathmax by abdo last…
Question Number 82456 by liki last updated on 21/Feb/20 Commented by abdomathmax last updated on 21/Feb/20 $${cos}\left(\mathrm{5}{x}\right)={sin}\left(\mathrm{4}{x}\right)\Leftrightarrow{cos}\left(\mathrm{5}{x}\right)={cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\right)\:\Rightarrow \\ $$$$\mathrm{5}{x}=\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\:+\mathrm{2}{k}\pi\:{or}\:\mathrm{5}{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{4}{x}\:+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$$\mathrm{9}{x}\:=\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:{or}\:{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\:\left({k}\in{Z}\right)\:\Rightarrow \\ $$$${x}=\frac{\pi}{\mathrm{18}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{9}}\:{or}\:{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi \\ $$$${S}\:=\left\{\frac{\pi}{\mathrm{18}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{9}}\:,{k}\epsilon{Z}\right\}\cup\left\{\left(\mathrm{2}{k}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi\:,{k}\in{Z}\right\}…
Question Number 16919 by rustylee last updated on 28/Jun/17 $${ax}\mathrm{2}+{bx}+{c}=\mathrm{0} \\ $$ Answered by RasheedSoomro last updated on 28/Jun/17 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\…
Question Number 147988 by puissant last updated on 24/Jul/21 $$\mathrm{La}\:\mathrm{somme}\:\mathrm{des}\:\mathrm{n}\:\mathrm{premiers}\:\mathrm{termes}\:\mathrm{d}'\mathrm{une} \\ $$$$\mathrm{s}\acute {\mathrm{e}rie}\:\mathrm{est}\:\mathrm{donn}\acute {\mathrm{e}}\:\mathrm{par}\:\mathrm{S}_{\mathrm{n}} =\mathrm{5n}^{\mathrm{2}} +\mathrm{2n}\:\mathrm{le}\: \\ $$$$\mathrm{n}−\mathrm{ieme}\:\mathrm{terme}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{serie}\:\mathrm{est}: \\ $$ Answered by Olaf_Thorendsen last updated…
Question Number 16918 by rustylee last updated on 28/Jun/17 $$\mathrm{2}{x}\mathrm{2}+\mathrm{9}{x}=\mathrm{10} \\ $$ Answered by Joel577 last updated on 28/Jun/17 $$\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:−\:\mathrm{10}\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} \:=\:\frac{−\mathrm{9}\:\pm\:\sqrt{\mathrm{81}\:+\:\mathrm{80}}}{\mathrm{4}} \\…
Question Number 82452 by mathmax by abdo last updated on 21/Feb/20 $${nature}\:{ofthe}\:{serie}\:\sum_{{n}=\mathrm{0}} ^{\infty} {ln}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\right) \\ $$ Commented by mathmax by abdo last updated on…
Question Number 16917 by tawa tawa last updated on 28/Jun/17 $$\mathrm{If}\:\:\:\mathrm{x}\:+\:\mathrm{iy}\:=\:\frac{\mathrm{1}}{\mathrm{a}\:+\:\mathrm{ib}} \\ $$$$\mathrm{prove}\:\mathrm{that}\::\:\:\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \right)\:=\:\mathrm{1} \\ $$ Answered by Tinkutara last updated on…
Question Number 82450 by M±th+et£s last updated on 21/Feb/20 $$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}+\left({tan}\left({x}\right)\right)^{\sqrt{\mathrm{2}}} }\:{dx} \\ $$ Commented by MJS last updated on 21/Feb/20 $$\mathrm{tan}^{\sqrt{\mathrm{2}}} \:{x}\:\notin\mathbb{R}\:\mathrm{for}\:\frac{\pi}{\mathrm{2}}<{x}<\pi \\…
Question Number 82448 by zainal tanjung last updated on 21/Feb/20 $$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{Lim}}\:\left\{\frac{\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)^{\mathrm{2}} −\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)\mathrm{sin}\:\left(\mathrm{4x}−\mathrm{2}\pi\right)}{\mathrm{5x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\mathrm{5x}−\frac{\mathrm{5}\pi}{\mathrm{2}}\:\right)}\right. \\ $$ Commented by jagoll last updated on 21/Feb/20 $${let}\:{x}−\frac{\pi}{\mathrm{2}}\:=\:{u} \\…
Question Number 82446 by M±th+et£s last updated on 21/Feb/20 $${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{−{log}\left({x}\right)} \:{x}\:{log}\left({x}\right)\:{dx}={e}\sqrt{\pi} \\ $$ Commented by abdomathmax last updated on 21/Feb/20 $${changement}\:{logx}={t}\:{give}\:{x}={e}^{{t}}…