Question Number 73538 by Rio Michael last updated on 13/Nov/19 $${find}\:{the}\:{general}\:{solution}\:{of}\: \\ $$$$\:{sin}\mathrm{4}{x}\:+\:{cos}\mathrm{2}{x}\:=\:\mathrm{0} \\ $$ Answered by ajfour last updated on 13/Nov/19 $$\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{2sin}\:\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:\:\:{or}…
Question Number 139075 by bramlexs22 last updated on 22/Apr/21 Answered by mr W last updated on 22/Apr/21 $$\mathrm{tan}\:\alpha=\frac{{x}}{\mathrm{144}} \\ $$$$\mathrm{tan}\:\beta=\frac{{x}}{\mathrm{144}+\mathrm{81}}=\frac{{x}}{\mathrm{225}} \\ $$$$\mathrm{tan}\:\gamma=\frac{{x}}{\mathrm{225}+\mathrm{99}}=\frac{{x}}{\mathrm{324}} \\ $$$$\mathrm{tan}\:\gamma=\mathrm{tan}\:\left(\mathrm{90}°−\alpha−\beta\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\beta\right)} \\…
Question Number 7996 by ridwan balatif last updated on 27/Sep/16 $${how}\:{to}\:{prove}\:{this} \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{10}^{{o}} }+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{20}^{{o}} }+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{40}}=\mathrm{12} \\ $$ Commented by prakash jain last…
Question Number 139024 by bramlexs22 last updated on 21/Apr/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}} \\ $$$$ \\ $$ Commented by EDWIN88 last updated on 21/Apr/21 $$\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$ Answered…
Question Number 138999 by bramlexs22 last updated on 21/Apr/21 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions} \\ $$$$\mathrm{of}\:\mathrm{equations}\:\sqrt{\mathrm{13}−\mathrm{18tan}\:\mathrm{x}}\:=\:\mathrm{6tan}\:\mathrm{x}−\mathrm{3} \\ $$$$\mathrm{where}\:−\mathrm{2}\pi<\mathrm{x}<\mathrm{2}\pi\:\mathrm{is} \\ $$ Answered by EDWIN88 last updated on 21/Apr/21 $$\left(\mathrm{1}\right)\:\mathrm{13}−\mathrm{18tan}\:\mathrm{x}\geqslant\mathrm{0}\:;\:\mathrm{tan}\:\mathrm{x}\leqslant\frac{\mathrm{13}}{\mathrm{18}} \\…
Question Number 7841 by ankursharma532 last updated on 19/Sep/16 $${tan}\Pi/\mathrm{16}=\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\:}\:−\:\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right) \\ $$ Commented by Yozzia last updated on 19/Sep/16 $${tan}\mathrm{4}{x}=\frac{\mathrm{2}{tan}\mathrm{2}{x}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{2}{x}} \\ $$$${Let}\:{x}=\pi/\mathrm{16}\Rightarrow{tan}\mathrm{4}{x}={tan}\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$$\therefore\mathrm{2}{tan}\mathrm{2}{x}=\mathrm{1}−{tan}^{\mathrm{2}}…
Question Number 7683 by new phone last updated on 08/Sep/16 $$\mathrm{If}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=\mathrm{2} \\ $$ Commented by sandy_suhendra last updated on 08/Sep/16 $${sin}\theta+^{} \mathrm{2}{cos}\theta=\mathrm{1} \\…
Question Number 7659 by Tawakalitu. last updated on 07/Sep/16 $${What}\:{is}\:{the}\:{sum}\:{of}\:{this}\:{sequence} \\ $$$${sin}^{\mathrm{2}} \left({x}\right)\:+\:{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\:{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\:+\:…\:+\:{sin}^{\mathrm{2}} \left({nx}\right) \\ $$ Commented by prakash jain last updated on…
Question Number 138725 by mathsuji last updated on 17/Apr/21 $${Solve}\:{for}\:{real}\:{numbers} \\ $$$$\frac{{sin}\left({sinx}\right)}{{sinx}}\:+\:\frac{{cos}\left({cosx}\right)}{{cosx}}\:=\:\mathrm{1} \\ $$ Answered by mr W last updated on 18/Apr/21 $${t}=\mathrm{sin}\:{x} \\ $$$$\mathrm{cos}\:{x}=\pm\sqrt{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 73160 by Aditya789 last updated on 07/Nov/19 $$\mathrm{1}+{tanAtan}\frac{{A}}{\mathrm{2}}={tanAcot}\frac{{A}}{\mathrm{2}}−\mathrm{1}={secA} \\ $$ Answered by $@ty@m123 last updated on 07/Nov/19 $$\mathrm{1}+\mathrm{tan}\:{A}\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\:.\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}^{\mathrm{2}}…