Question Number 157080 by amin96 last updated on 19/Oct/21 Commented by cortano last updated on 20/Oct/21 $$\Rightarrow\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}=\frac{\mathrm{3sin}\:\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}−\mathrm{cos}\:\mathrm{2}{x}=\mathrm{3sin}\:\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{3}−\mathrm{cos}\:\mathrm{2}{x}=\mathrm{3}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)} \\ $$$$\Rightarrow\mathrm{9}+\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{6cos}\:\left(\mathrm{2}{x}\right)=\mathrm{9}−\mathrm{9cos}\:^{\mathrm{2}}…
Question Number 157071 by MathSh last updated on 19/Oct/21 $$\mathrm{if}\:\:\:\mathrm{0}<\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\boldsymbol{\mathrm{x}}}\:+\:\frac{\mathrm{2}}{\pi\mathrm{x}}\:\leqslant\:\left(\mathrm{1}\:-\:\frac{\mathrm{2}}{\pi}\right)^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{2}}{\pi} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157057 by amin96 last updated on 19/Oct/21 $$\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{13}}+\frac{\mathrm{1}}{\mathrm{19}}={a} \\ $$$${find}\:\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{13}}+\frac{\mathrm{6}}{\mathrm{19}}=? \\ $$ Commented by aliyn last updated on 19/Oct/21 $$\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{13}}+\frac{\mathrm{6}}{\mathrm{19}}=\left(\frac{\mathrm{2}}{\mathrm{7}}\:+\:\frac{\mathrm{2}}{\mathrm{13}}\:+\frac{\mathrm{2}}{\mathrm{19}}\:\right)+\left(\frac{\mathrm{2}}{\mathrm{13}}+\frac{\mathrm{4}}{\mathrm{19}}\right) \\ $$$$ \\…
Question Number 157053 by depressiveshrek last updated on 19/Oct/21 $$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{write}\:\mathrm{any}\:\mathrm{quadratic}\:\mathrm{function}\:\mathrm{in}\:\mathrm{vertex}\:\mathrm{form},\:\mathrm{just}\:\mathrm{use}\:\mathrm{these}\:\mathrm{formulas}: \\ $$$$\: \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\: \\ $$$$\mathrm{if}\:{a}>\mathrm{0},\:\mathrm{then}: \\ $$$${f}\left({x}\right)=\left({x}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{c} \\ $$$$\:…
Question Number 157044 by MathSh last updated on 18/Oct/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\left(\mathrm{sin2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{4cos}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:+\:\mathrm{1}\right)\left(\mathrm{cos5}\boldsymbol{\mathrm{x}}\:-\:\mathrm{cos}\boldsymbol{\mathrm{x}}\right)<\mathrm{0} \\ $$ Answered by mindispower last updated on 19/Oct/21 $$\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)=\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}}…
Question Number 157045 by MathSh last updated on 18/Oct/21 Commented by Tawa11 last updated on 19/Oct/21 $$\mathrm{Great}\:\mathrm{sir} \\ $$ Answered by mr W last updated…
Question Number 25971 by Tinkutara last updated on 17/Dec/17 $${In}\:{finding}\:{the}\:{equations}\:{of}\:{the} \\ $$$${bisectors}\:{of}\:{the}\:{angles}\:{between}\:{two} \\ $$$${lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}, \\ $$$${why}\:{we}\:{observe}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}}…
Question Number 157035 by Armindo last updated on 18/Oct/21 Commented by Armindo last updated on 18/Oct/21 $${help}… \\ $$ Commented by MJS_new last updated on…
Question Number 25961 by kaivan.ahmadi last updated on 16/Dec/17 $$\mathrm{8cos}^{\mathrm{4}} \mathrm{x}−\mathrm{8cos}^{\mathrm{2}} \mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}:\mathrm{8cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$−\mathrm{8cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sinx}=\underset{−} {+}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow \\…
Question Number 157021 by MathSh last updated on 18/Oct/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\left(\boldsymbol{\mathrm{n}}\right)\:=\underset{\:\mathrm{1}} {\overset{\:\boldsymbol{\mathrm{n}}} {\int}}\left(\left[\mathrm{x}\right]^{\mathrm{2}} \centerdot\left\{\mathrm{x}\right\}\:+\:\left[\mathrm{x}\right]\centerdot\left\{\mathrm{x}\right\}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{n}\in\mathbb{N}\:\:;\:\:\left[\ast\right]-\mathrm{GIF}\:\:;\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$ Answered by ghimisi last updated…