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?\\n | ||
Commented bymr W last updated on 17/Mar/21 | ||
$$\mathrm{2}\:\mathrm{cos}^{\mathrm{3}} \:{x}+\mathrm{sin}\:{x}−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$ $$\mathrm{2}\:\mathrm{cos}^{\mathrm{3}} \:{x}+\mathrm{sin}\:{x}−\mathrm{3}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}\right)\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$ $$\mathrm{5}\:\mathrm{cos}^{\mathrm{3}} \:{x}+\mathrm{sin}\:{x}−\mathrm{3}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$ $$\mathrm{cos}\:{x}\:\left(\mathrm{5}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{3}\right)+\mathrm{sin}\:{x}=\mathrm{0} \\ $$ $$\mathrm{3}−\frac{\mathrm{5}}{\mathrm{tan}^{\mathrm{2}} \:{x}+\mathrm{1}}=\mathrm{tan}\:{x} \\ $$ $$\mathrm{3}−\frac{\mathrm{5}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{t} \\ $$ $${t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +{t}+\mathrm{2}=\mathrm{0} \\ $$ $$\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)=\mathrm{0} \\ $$ $$\Rightarrow{t}=\mathrm{tan}\:{x}=\mathrm{2}\:\Rightarrow{\color{mathred}{x}}\color{mathred}{=}{\color{mathred}{n}}\color{mathred}{\pi}\color{mathred}{+}\mathrm{\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}}^{\color{mathred}{−}\mathrm{\color{mathred}{1}}} \mathrm{\color{mathred}{2}} \\ $$ $$\Rightarrow{t}=\mathrm{tan}\:{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow{\color{mathred}{x}}\color{mathred}{=}{\color{mathred}{n}}\color{mathred}{\pi}\color{mathred}{+}\mathrm{\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}}^{\color{mathred}{−}\mathrm{\color{mathred}{1}}} \frac{\mathrm{\color{mathred}{1}}\color{mathred}{\pm}\sqrt{\mathrm{\color{mathred}{5}}}}{\mathrm{\color{mathred}{2}}} \\ $$ | ||
Commented byliberty last updated on 17/Mar/21 | ||
$${replacing}\:\mathrm{sin}\:{x}=\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:{x}\:\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$ $$\left(\bullet\right)\:\mathrm{2cos}\:^{\mathrm{3}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:{x}\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$ $${divided}\:{by}\:\mathrm{cos}\:^{\mathrm{3}} {x} \\ $$ $$\mathrm{\color{mathblue}{2}}\color{mathblue}{+}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}^{\mathrm{\color{mathblue}{3}}} {\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}{\color{mathblue}{x}}\color{mathblue}{−}\mathrm{\color{mathblue}{3}\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}^{\mathrm{\color{mathblue}{2}}} {\color{mathblue}{x}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\mathrm{\color{mathblue}{0}}\color{mathblue}{\:} \\ $$ $$\color{mathblue}{\Rightarrow}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}^{\mathrm{\color{mathblue}{3}}} {\color{mathblue}{x}}\color{mathblue}{−}\mathrm{\color{mathblue}{3}\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}^{\mathrm{\color{mathblue}{2}}} {\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}{\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{2}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\mathrm{\color{mathblue}{0}} \\ $$ $${\color{mathblue}{l}\color{mathblue}{e}\color{mathblue}{t}}\color{mathblue}{\:}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}{\color{mathblue}{x}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}{\color{mathblue}{u}} \\ $$ $$\color{mathblue}{\Rightarrow}{\color{mathblue}{u}}^{\mathrm{\color{mathblue}{3}}} \color{mathblue}{−}\mathrm{\color{mathblue}{3}}{\color{mathblue}{u}}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{+}{\color{mathblue}{u}}\color{mathblue}{+}\mathrm{\color{mathblue}{2}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\mathrm{\color{mathblue}{0}} \\ $$ $$\color{mathblue}{\left(}{\color{mathblue}{u}}\color{mathblue}{−}\mathrm{\color{mathblue}{2}}\color{mathblue}{\right)}\color{mathblue}{\left(}{\color{mathblue}{u}}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{−}{\color{mathblue}{u}}\color{mathblue}{−}\mathrm{\color{mathblue}{1}}\color{mathblue}{\right)}\color{mathblue}{=}\mathrm{\color{mathblue}{0}} \\ $$ $$\color{mathblue}{\rightarrow}\begin{cases}{{\color{mathblue}{u}}\color{mathblue}{=}\mathrm{\color{mathblue}{2}}\color{mathblue}{\Rightarrow}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}{\color{mathblue}{x}}\color{mathblue}{=}\mathrm{\color{mathblue}{2}}}\\{{\color{mathblue}{u}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\frac{\mathrm{\color{mathblue}{1}}\color{mathblue}{\pm}\sqrt{\mathrm{\color{mathblue}{5}}}}{\mathrm{\color{mathblue}{2}}}\color{mathblue}{\Rightarrow}\mathrm{\color{mathblue}{t}\color{mathblue}{a}\color{mathblue}{n}}\color{mathblue}{\:}{\color{mathblue}{x}}\color{mathblue}{=}\frac{\mathrm{\color{mathblue}{1}}\color{mathblue}{\pm}\sqrt{\mathrm{\color{mathblue}{5}}}}{\mathrm{\color{mathblue}{2}}}}\end{cases} \\ $$ | ||
Answered by MJS_new last updated on 17/Mar/21 | ||
$$\mathrm{let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$ $$−\frac{\mathrm{2}\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right)\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} −\mathrm{6}{t}^{\mathrm{3}} +\mathrm{2}{g}+\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0} \\ $$ $$\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){t}−\mathrm{1}\right)=\mathrm{0} \\ $$ $$\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{t}\:\mathrm{and}\:\mathrm{then}\:{x}=\mathrm{2}\pi{n}+\mathrm{2arctan}\:{t} \\ $$ | ||