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Author: Tinku Tara

tan-2-8sin-5cos-sin-3-cos-3-cos-

Question Number 194767 by horsebrand11 last updated on 15/Jul/23 $$\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2}\: \\ $$$$\:\:\:\frac{\mathrm{8sin}\:\theta+\mathrm{5cos}\:\theta}{\mathrm{sin}\:^{\mathrm{3}} \theta+\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{cos}\:\theta}\:=? \\ $$ Answered by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{sec}\:^{\mathrm{2}} {x}=\mathrm{5}\:,\:\mathrm{tan}\:^{\mathrm{2}}…

1-2cot-2x-cot-x-3-x-

Question Number 194766 by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{1}+\mathrm{2cot}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\:\mathrm{3}\: \\ $$$$\:\:\:{x}=? \\ $$ Answered by Frix last updated on 15/Jul/23 $$\mathrm{1}+\mathrm{2cor}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\mathrm{3} \\ $$$$\mathrm{cot}^{\mathrm{2}} \:{x}\:=\mathrm{3}…

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Question Number 194756 by horsebrand11 last updated on 15/Jul/23 $$\:\:\:\:\:\:\underbrace{\boldsymbol{{b}}} \\ $$ Answered by cortano12 last updated on 15/Jul/23 $$\:\:\frac{\mathrm{x}−\mathrm{a}}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}}\:=\:\frac{\mathrm{x}−\mathrm{a}}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:+\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\frac{\mathrm{3}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}−\frac{\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\:\frac{\mathrm{2}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\…

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Question Number 194791 by dimentri last updated on 15/Jul/23 $$\:\:\:\underline{\underbrace{\boldsymbol{{x}}}} \\ $$ Answered by Frix last updated on 15/Jul/23 $$\mathrm{If}\:\sqrt[{\mathrm{7}}]{−{r}}=−\sqrt[{\mathrm{7}}]{{r}} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} }\sqrt[{\mathrm{7}}]{{x}−\sqrt{\mathrm{2}}}=\frac{{x}^{\frac{\mathrm{9}}{\mathrm{7}}} }{\mathrm{2}\sqrt[{\mathrm{7}}]{{x}+\sqrt{\mathrm{2}}}}…

Question-194790

Question Number 194790 by BagusSetyoWibowo last updated on 15/Jul/23 Answered by Frix last updated on 15/Jul/23 $$\mathrm{Where}'\mathrm{s}\:\mathrm{the}\:\mathrm{problem}? \\ $$$${ab}^{{x}+{c}} ={d}\:\Rightarrow\:{x}=\frac{\mathrm{ln}\:\frac{{d}}{{a}}}{\mathrm{ln}\:{b}}−{c} \\ $$$${x}=\frac{\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{54}\:\left(\mathrm{log}_{\mathrm{5}} \:\mathrm{60}\:+\frac{\tau}{\mathrm{60}}+\mathrm{sin}\:\left(\mathrm{8}+\mathrm{cot}\:\mathrm{67}\right)\:+\mathrm{4}^{\mathrm{2}} \right)}{\mathrm{2}}}{\mathrm{ln}\:\mathrm{50}}−\pi \\…