Question Number 18457 by Tinkutara last updated on 21/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interval}\:\left[\mathrm{0},\:\mathrm{4}\pi\right]\:\mathrm{is} \\ $$ Answered by mrW1 last updated on 21/Jul/17 $$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\…
Question Number 18455 by Tinkutara last updated on 21/Jul/17 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+ \\ $$$$\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\:=\:\mathrm{6},\:\mathrm{where}\:\mathrm{0}\:<\:{x},\:{y},\:{z}\:<\:\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$$${x}\:+\:{y}\:+\:{z}\:=\:\pi,\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Three}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Two}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{solutions} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Just}\:\mathrm{one}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\…
Question Number 18456 by Tinkutara last updated on 21/Jul/17 $$\mathrm{The}\:\mathrm{complete}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:−\:\mathrm{12}\left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)\:+\:\mathrm{12}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)\frac{\pi}{\mathrm{4}},\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{4}\right)\:{x}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$ Answered…
Question Number 83975 by jagoll last updated on 08/Mar/20 $$\mathrm{find}\:\mathrm{solution} \\ $$$$\mathrm{8}\:\mathrm{tan}\:\mathrm{x}−\mathrm{8tan}\:^{\mathrm{5}} \mathrm{x}\:=\:\mathrm{sec}\:^{\mathrm{6}} \mathrm{x}\:\mathrm{in}\:\mathrm{x}\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right) \\ $$ Answered by TANMAY PANACEA last updated on 08/Mar/20 $$\mathrm{8}{a}−\mathrm{8}{a}^{\mathrm{5}}…
Question Number 18426 by Tinkutara last updated on 20/Jul/17 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{2}\:\mathrm{cot}\:\mathrm{2}{x}\:−\:\mathrm{3}\:\mathrm{cot}\:\mathrm{3}{x}\:=\:\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Two}\:\mathrm{solutions}\:\mathrm{in}\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{One}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{No}\:\mathrm{solution}\:\mathrm{in}\:\left(−\infty,\:\infty\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Three}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{0},\:\pi\right) \\ $$ Terms of Service…
Question Number 18402 by Tinkutara last updated on 20/Jul/17 $$\mathrm{If}\:{P}_{{n}} \:=\:\mathrm{cos}^{{n}} \:\theta\:+\:\mathrm{sin}^{{n}} \:\theta,\:\theta\:\in\:\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right],\:{n}\:\in \\ $$$$\left(−\infty,\:\mathrm{2}\right),\:\mathrm{then}\:\mathrm{minimum}\:\mathrm{of}\:{P}_{{n}} \:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 149457 by liberty last updated on 05/Aug/21 Answered by EDWIN88 last updated on 05/Aug/21 $$\:{f}\left({x}\right)=\mathrm{sin}\:^{\mathrm{4}} {x}−\mathrm{sin}\:{x}\mathrm{cos}\:{x}+\mathrm{cos}\:^{\mathrm{4}} {x} \\ $$$${f}\left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x} \\ $$$${we}\:{define}\:\mathrm{sin}\:\mathrm{2}{x}={t} \\…
Question Number 149447 by liberty last updated on 05/Aug/21 $$\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\: \\ $$$$\:\begin{cases}{\mathrm{sin}\:\mathrm{2x}+\mathrm{cos}\:\mathrm{3y}=−\mathrm{1}}\\{\sqrt{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{y}}\:+\sqrt{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}\:=\mathrm{1}+\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)}\end{cases} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 83864 by jagoll last updated on 07/Mar/20 $$\mathrm{what}\:\mathrm{Maclaurin}\:\mathrm{series}\:\mathrm{of}\:\mathrm{function} \\ $$$$\mathrm{tan}\:\left(\mathrm{x}\right)? \\ $$ Commented by niroj last updated on 07/Mar/20 $$\:\mathrm{Solution}: \\ $$$$\:\:\mathrm{let},\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{tan}\:\mathrm{x}\:\:,\:\:\mathrm{f}_{\mathrm{0}} \left(\mathrm{0}\right)=\mathrm{0}…
Question Number 149385 by liberty last updated on 05/Aug/21 Answered by Ar Brandon last updated on 05/Aug/21 $$\alpha+\frac{\mathrm{1}}{\alpha}=\mathrm{3}\Rightarrow\alpha^{\mathrm{2}} −\mathrm{3}\alpha+\mathrm{1}=\mathrm{0}\Rightarrow\alpha=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{tan}{x}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow\mathrm{tan}{x}=\frac{\mathrm{14}\pm\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{3}} {x}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{3}} {x}}=\left(\frac{\mathrm{14}\pm\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{3}}…