Question Number 98191 by behi83417@gmail.com last updated on 12/Jun/20 $$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\::\left\{\boldsymbol{\mathrm{x}}\mid\:\mathrm{0}<\boldsymbol{\mathrm{x}}<\mathrm{2}\boldsymbol{\pi}\right\}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{cosec}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}−\mathrm{9}=\boldsymbol{\mathrm{cotx}} \\ $$ Answered by MJS last updated on 12/Jun/20 $$\mathrm{4cosec}^{\mathrm{2}} \:{x}\:−\mathrm{9}=\mathrm{cot}\:{x} \\…
Question Number 98149 by me2love2math last updated on 11/Jun/20 Commented by me2love2math last updated on 11/Jun/20 $${Q}\mathrm{1}\:{and}\:\mathrm{2}\:{pls} \\ $$ Answered by mr W last updated…
Question Number 32538 by GAUTHAM last updated on 27/Mar/18 Commented by prof Abdo imad last updated on 28/Mar/18 $${we}\:{have}\:{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)=\:\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow\:{sin}\left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:\:\Rightarrow \\ $$$${arcsin}\left(\:\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{12}}\:. \\…
Question Number 97994 by hardylanes last updated on 10/Jun/20 $${find}\:{all}\:{the}\:{values}\:{of}\:\theta,\:{in}\:{the}\:{interval}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$${for}\:{which}\:\mathrm{sin}\:\mathrm{3}\theta−\mathrm{sin}\:\theta=\sqrt{\mathrm{3cos}\:\mathrm{2}\theta} \\ $$ Answered by Rio Michael last updated on 10/Jun/20 $$\mathrm{i}\:\mathrm{pressume}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\theta\:−\:\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{for}\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\mathrm{2}\pi…
Question Number 32163 by jarjum last updated on 20/Mar/18 $${If}\:{the}\:{corrdinater}\:{of}\:{the}\:{verticle}\:{of}\:{an} \\ $$$${eqvilateral}\:{triangle}\:{with}\:{length}\:{x}\:{are} \\ $$$$\left({x}_{\mathrm{1}+} {y}_{\mathrm{1}} \right),\left({y}_{\mathrm{1}} +{y}_{\mathrm{2}} \right)\:{and}\:\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right)\:{then} \\ $$$$\left(\begin{vmatrix}{{x}_{\mathrm{1}} \:\:\:{y}_{\mathrm{1}} \:\:\:\mathrm{2}}\\{{x}_{\mathrm{2}} \:\:\:{y}_{\mathrm{2}}…
Question Number 32164 by jarjum last updated on 20/Mar/18 $${Determine}\:{wether}\:{the}\:{relation}\:{on}\:{the} \\ $$$${set}\:{R}\:{of}\:{all}\:{real}\:{number}\:{as} \\ $$$${R}=\left\{\left({a},{b}\right):{a},{b}\in{R}\:{and}\:{a}−{b}\:+\sqrt{\mathrm{3}}\:\in\:{s}\:\right. \\ $$$$\left.{where}\:{s}\:{is}\:{the}\:{set}\:{of}\:{all}\:{irrational}\:{no}\right) \\ $$$${is}\:{reflexive},\:{symmetric}\:{and}\:{transitive}? \\ $$ Terms of Service Privacy Policy…
Question Number 32161 by jarjum last updated on 20/Mar/18 $${Let}\:{a}\:{function}\:{F}\::{R}\rightarrow{R}\:{be}\:{defined}\:{by} \\ $$$${f}\left({x}\right)=\mathrm{1}+{ax},\alpha\neq\:\mathrm{0}\:{for}\:{all}\:{X}\:\in\:{R}.\:{Show} \\ $$$${that}\:{f}\:{is}\:{invertible}\:{and}\:{find}\:{its}\:{inverse} \\ $$$${function}.{Also}\:{find}\:{the}\:{value}\:\left({s}\right)\:{of}\:\alpha \\ $$$${if}\:{inverse}\:{of}\:{f}\:{is}\:{itself} \\ $$ Answered by mrW2 last updated…
Question Number 32156 by jarjum last updated on 20/Mar/18 $${evaluate}\int\frac{\sqrt{{cos}\:\mathrm{2}{x}}}{{sin}\:{x}}{dx} \\ $$ Commented by abdo imad last updated on 20/Mar/18 $${I}\:=\:\int\:\sqrt{\frac{{cos}\left(\mathrm{2}{x}\right)}{{sin}^{\mathrm{2}} {x}}}\:{dx}\:=\int\:\sqrt{\frac{{cos}\left(\mathrm{2}{x}\right)}{\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}}\:{dx} \\ $$$$=\int\:\:\sqrt{\:\:\:\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}}\:{dx}\:\:{ch}.\:{tanx}={t}\:{give} \\…
Question Number 97680 by bemath last updated on 09/Jun/20 Commented by bobhans last updated on 09/Jun/20 $$\mathrm{cos}\:\mathrm{6A}−\mathrm{cos}\:\mathrm{2A}=−\mathrm{2sin}\:\mathrm{4Asin}\:\mathrm{2A} \\ $$$$\Leftrightarrow−\mathrm{2sin}\:\mathrm{4Asin}\:\mathrm{2A}−\mathrm{2cos}\:\mathrm{4A}+\mathrm{2}= \\ $$$$\Leftrightarrow\:−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{2A}\:\mathrm{cos}\:\mathrm{2A}−\mathrm{2}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2A}\right)+\mathrm{2} \\ $$$$\Leftrightarrow−\mathrm{4sin}\:^{\mathrm{2}}…
Question Number 97546 by bobhans last updated on 08/Jun/20 Commented by bobhans last updated on 08/Jun/20 can anyone translate to english? Commented by bemath last updated on 08/Jun/20 $$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:…