Question Number 135791 by benjo_mathlover last updated on 16/Mar/21
$${If}\:{x}^{\mathrm{2}} +\left(\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\right){x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$${has}\:{two}\:{real}\:{solutions}\: \\ $$$$\left\{\:\mathrm{2}−\sqrt{\mathrm{3}}\:,\:\mathrm{2}+\sqrt{\mathrm{3}}\:\right\},\:{find}\:\begin{cases}{\mathrm{sin}\:\theta}\\{\mathrm{cos}\:\theta}\end{cases}\:. \\ $$
Answered by EDWIN88 last updated on 17/Mar/21
$$\mathrm{By}\:\mathrm{Vieta}'\mathrm{s}\:\Rightarrow\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \:=\:−\frac{\mathrm{b}}{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\:=\:−\mathrm{4}\:,\:\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=−\mathrm{4} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{4tan}\:\theta+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{tan}\:\theta+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\:=\:\mathrm{0}\:\Rightarrow\mathrm{tan}\:\theta\:=\:−\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\mathrm{or}\:\mathrm{tan}\:^{\mathrm{2}} \theta\:=\:\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}\:\rightarrow\begin{cases}{\mathrm{sec}\:^{\mathrm{2}} \theta\:=\:\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}}\\{\mathrm{cos}\:^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{cos}\:\theta\:=\:\pm\sqrt{\frac{\mathrm{1}}{\left(\sqrt{\mathrm{6}}\:\pm\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }}\:=\:\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\pm\sqrt{\mathrm{2}}}}\\{\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}}}\end{cases} \\ $$$$ \\ $$