Question Number 207562 by hardmath last updated on 18/May/24 $$\mathrm{4}\:\mathrm{cos}\:\mathrm{50}°\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Commented by Frix last updated on 18/May/24 $$−\mathrm{2tan}\:\mathrm{10}° \\ $$ Terms of Service…
Question Number 207561 by hardmath last updated on 18/May/24 $$\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{2} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by mathzup last updated on 18/May/24 $${e}\Leftrightarrow\mathrm{4}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\:+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow \\ $$$$\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow…
Question Number 207558 by hardmath last updated on 18/May/24 $$\left(\mathrm{x}\:−\:\mathrm{2}\right)\:\mathrm{lg}\:\frac{\mathrm{x}}{\mathrm{3}}\:\:\geqslant\:\:\mathrm{0} \\ $$ Answered by mr W last updated on 19/May/24 $${x}−\mathrm{2}\geqslant\mathrm{0}\:\wedge\:\frac{{x}}{\mathrm{3}}\geqslant\mathrm{1}\:\Rightarrow{x}\geqslant\mathrm{3}\:\checkmark \\ $$$${or} \\ $$$${x}−\mathrm{2}\leqslant\mathrm{0}\:\wedge\:\mathrm{0}<\frac{{x}}{\mathrm{3}}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}<{x}\leqslant\mathrm{2}\:\checkmark…
Question Number 207563 by hardmath last updated on 18/May/24 $$\mathrm{z}\:\:+\:\:\mid\mathrm{z}\mid\:\:=\:\:\mathrm{1}\:\:+\:\:\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\varphi}\:=\:? \\ $$ Answered by Frix last updated on 18/May/24 $${z}+\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}−{z} \\…
Question Number 207559 by hardmath last updated on 18/May/24 $$\mathrm{log}_{\boldsymbol{\mathrm{x}}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7}\right)\:\:\leqslant\:\:\mathrm{1} \\ $$ Answered by mr W last updated on 19/May/24 $$\mathrm{log}_{{x}} \:\left({x}^{\mathrm{2}} +\mathrm{7}\right)−\mathrm{log}_{{x}}…
Question Number 207518 by hardmath last updated on 17/May/24 $$\mathrm{cos}\boldsymbol{\mathrm{x}}\:\mathrm{cos3}\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{cos5}\boldsymbol{\mathrm{x}}\:\mathrm{cos7}\boldsymbol{\mathrm{x}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 17/May/24 $$\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{7}{x} \\ $$$$\frac{\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{12}{x}}{\mathrm{2}} \\…
Question Number 207519 by hardmath last updated on 17/May/24 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$ Answered by Frix last updated on 17/May/24 $${f}\left({x}\right)=\frac{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{\mid{x}+\mathrm{2}\mid}=\begin{cases}{−\left({x}+\mathrm{1}\right),\:{x}<−\mathrm{2}}\\{\left({x}+\mathrm{1}\right),\:−\mathrm{2}<{x}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{approach}\:\mathrm{2}\:\mathrm{from}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{side}\:\mathrm{but} \\…
Question Number 207486 by hardmath last updated on 17/May/24 Commented by som(math1967) last updated on 17/May/24 $${DE}\bot{AK}\:? \\ $$ Commented by hardmath last updated on…
Question Number 207487 by hardmath last updated on 17/May/24 $$\mid\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:−\:\mathrm{4}\mid\:=\:\mid\mathrm{x}\:−\:\mathrm{4}\mid \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{max}}\:\:=\:\:? \\ $$ Answered by A5T last updated on 17/May/24 $$\mid{x}−\mathrm{4}\mid\mid{x}+\mathrm{1}\mid=\mid{x}−\mathrm{4}\mid\Rightarrow\mid{x}−\mathrm{4}\mid=\mathrm{0}\:{or}\:\mid{x}+\mathrm{1}\mid=\mathrm{1} \\ $$$$\Rightarrow{x}−\mathrm{4}=\mathrm{0}\:{or}\:{x}+\mathrm{1}=\mathrm{1}\:{or}\:{x}+\mathrm{1}=−\mathrm{1}…
Question Number 207498 by hardmath last updated on 17/May/24 $$\mathrm{cos2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:=\:\mathrm{tg}\left(\mathrm{225}°\right)\centerdot\left(\mathrm{0},\mathrm{360}°\right) \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}\:=\:? \\ $$ Answered by MM42 last updated on 17/May/24 $$−\mathrm{2}{sin}^{\mathrm{2}} {x}+{sinx}=\mathrm{0} \\ $$$${sinx}=\mathrm{0}\Rightarrow{x}=\pi…