Question Number 136268 by EDWIN88 last updated on 20/Mar/21 $$\mathrm{Solve}\:\mathrm{sin}\:\left(\mathrm{2x}\right)−\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{for}\: \\ $$$$\mathrm{x}\:\mathrm{real}\:\mathrm{and}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 70597 by Maclaurin Stickker last updated on 06/Oct/19 $${solve} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$ Answered by Kunal12588 last updated on 08/Oct/19 $${cos}\:\mathrm{2}\alpha\:=\:\mathrm{2}{cos}^{\mathrm{2}}…
Question Number 136121 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{13}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:+\:\mathrm{17}\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3cos}\:\mathrm{3t}\:=\:\mathrm{13sin}\:\mathrm{2t}+\mathrm{17cos}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{3cos}\:\mathrm{t}\right)=\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:+\mathrm{17cos}\:\mathrm{t} \\ $$$$\mathrm{12cos}\:^{\mathrm{3}}…
Question Number 136122 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}\right)\:+\mathrm{sin}\:\mathrm{2}{x}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)=\mathrm{2sin}\:{x}\: \\ $$ Answered by EDWIN88 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{xcos}\:\mathrm{x}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{x}\right)−\mathrm{2sin}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{x}\left[\:\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\:\right]=\mathrm{0}…
Question Number 136094 by mnjuly1970 last updated on 19/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:….\:{nice}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:::\:{cos}\left({A}\right){cos}\left({B}\right){cos}\left({C}\right)=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\:\:\:{find}\:\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({C}\right)=? \\ $$ Commented by MJS_new…
Question Number 5008 by Rasheed Soomro last updated on 31/Mar/16 $$\mathrm{A}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{plastic}\:\mathrm{strip}\:\mathrm{1}\:\mathrm{metre}\:\mathrm{long} \\ $$$$\mathrm{is}\:\mathrm{bent}\:\mathrm{to}\:\mathrm{form}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{95}°\:\mathrm{as}\:\mathrm{measure}\:\mathrm{of}\:\mathrm{its}\:\mathrm{largest}\:\mathrm{angle}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}. \\ $$ Commented by prakash jain last updated…
Question Number 136008 by liberty last updated on 17/Mar/21 $$\:\mathrm{cos}\:{x}\:\sqrt{\mathrm{tan}\:{x}}\:=\:\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x} \\ $$ Answered by MJS_new last updated on 17/Mar/21 $${t}=\sqrt{\mathrm{tan}\:{x}}\:\wedge{t}\geqslant\mathrm{0} \\ $$$$\frac{{t}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}}…
Question Number 135974 by liberty last updated on 17/Mar/21 $$ \\ $$How do you solve for 2cos^3(x) + sinx – 3sin^2 (x) cos(x) =0? Commented…
Question Number 70394 by Cmr 237 last updated on 04/Oct/19 $$\mathrm{montrer}\:\mathrm{que} \\ $$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=\mathrm{4cos}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{C}}{\mathrm{2}} \\ $$ Answered by $@ty@m123 last updated on 05/Oct/19 $${LHS}=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:{C} \\ $$$$=\mathrm{2sin}\:\frac{\pi−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:{C}…
Question Number 135926 by liberty last updated on 17/Mar/21 $$ \\ $$If cos(A+B) =1/2, sinA=1/√2, then what is cos (A – B)? Answered by benjo_mathlover…