Question Number 80519 by mathocean1 last updated on 03/Feb/20 $$\mathrm{g}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} {x}+{sin}\left(\mathrm{2}{x}\right). \\ $$$${g}'\left({x}\right)=\:……….? \\ $$ Commented by mr W last updated on 03/Feb/20 $${g}'\left({x}\right)=\mathrm{2}×\mathrm{2}\:\mathrm{cos}\:{x}\:\left(−\mathrm{sin}\:{x}\right)+\mathrm{cos}\:\left(\mathrm{2}{x}\right)\:\mathrm{2} \\…
Question Number 14977 by Tinkutara last updated on 06/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{7}} \:{x}\:=\:\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[−\pi,\:\pi\right]. \\ $$ Commented by mrW1 last updated on 06/Jun/17 $$\mathrm{cos}\:{x}=\pm\mathrm{1}\:{and}\:\mathrm{sin}\:{x}=\mathrm{0} \\…
Question Number 145979 by iloveisrael last updated on 09/Jul/21 $$\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{1}=\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{2cos}\:\mathrm{x}−\mathrm{2cos}\:\mathrm{xsin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{x}=? \\ $$ Answered by Olaf_Thorendsen last updated on 10/Jul/21…
Question Number 14878 by Tinkutara last updated on 05/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}\:\mathrm{sin}\:{x}\:=\:\mathrm{2} \\ $$ Commented by ajfour last updated on 05/Jun/17 $${in}\:{such}\:{a}\:{question}\:{interval}\:{is}\: \\ $$$${mentioned},\:{or}\:{we}\:{have}\:{infinite} \\…
Question Number 14879 by Tinkutara last updated on 05/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x},\:\mathrm{satisfying} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{sin}\:{x}\:−\:\mathrm{2}\:\geqslant\:\mathrm{0} \\ $$ Answered by ajfour last updated on 05/Jun/17 $$\left(\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\…
Question Number 14758 by Tinkutara last updated on 04/Jun/17 $$\mathrm{Solve}\:\mathrm{tan}\:{x}\:+\:\mathrm{tan}\:\mathrm{2}{x}\:+\:\mathrm{tan}\:\mathrm{3}{x}\:=\:\mathrm{0} \\ $$ Commented by Tinkutara last updated on 04/Jun/17 $$\mathrm{Answer}:\:{x}\:=\:\frac{{n}\pi}{\mathrm{3}}\:,\:{n}\pi\:\pm\:\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{where}\:{n}\:=\:\mathrm{0},\:\pm\:\mathrm{1},\:\pm\:\mathrm{2} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{given}\:\mathrm{exactly}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}.…
Question Number 14752 by Tinkutara last updated on 04/Jun/17 $$\mathrm{Find}\:\mathrm{ordered}\:\mathrm{pair}\:\mathrm{of}\:\left({x},\:{y}\right)\:\mathrm{given}\:{x},\:{y}\:\in \\ $$$$\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{if}\:\mathrm{3}^{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{y}} \:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{5}^{\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:{y}} \:=\:\mathrm{5} \\ $$ Commented by Tinkutara last updated…
Question Number 145814 by gsk2684 last updated on 08/Jul/21 $${Let}\:{f}\left({x}\right)=\left\{\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)+\mathrm{sin}\:\left(\mathrm{cot}^{−\mathrm{1}} {x}\right)\right\}^{\mathrm{2}} −\mathrm{1} \\ $$$${where}\:\mid{x}\mid>\mathrm{1}\:{and}\:\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\frac{{d}}{{dx}}\left(\mathrm{sin}^{−\mathrm{1}} {f}\left({x}\right)\right). \\ $$$${if}\:{y}\left(\sqrt{\mathrm{3}}\right)=\frac{\Pi}{\mathrm{6}}\:{then}\:{y}\left(−\sqrt{\mathrm{3}}\right)=? \\ $$ Terms of Service Privacy Policy…
Question Number 14633 by Tinkutara last updated on 03/Jun/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\left(\mathrm{1}\:−\:\mathrm{tan}\theta\right)\left(\mathrm{1}\:+\:\mathrm{sin2}\theta\right)\:=\:\mathrm{1}\:+\:\mathrm{tan}\theta \\ $$ Commented by myintkhaing last updated on 03/Jun/17 $$\frac{\mathrm{1}−{tan}\theta}{\mathrm{1}+{tan}\theta}\:\left(\mathrm{1}+{sin}\mathrm{2}\theta\right)\:=\:\mathrm{1} \\ $$$$\frac{{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta}\:\left(\mathrm{1}+{sin}\mathrm{2}\theta\right)\:=\:\mathrm{1} \\…
Question Number 14632 by Tinkutara last updated on 03/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\mathrm{3}^{\mathrm{sin}\:\mathrm{2}{x}\:+\:\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{x}} \:+\:\mathrm{3}^{\mathrm{1}\:−\:\mathrm{sin}\:\mathrm{2}{x}\:+\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}} \:=\:\mathrm{28} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated…