Question Number 127814 by dactor last updated on 02/Jan/21 | ||
Answered by mnjuly1970 last updated on 02/Jan/21 | ||
$${solution} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}−{cos}\left({x}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{cos}\left({x}\right)\left(\mathrm{2}{cos}\left({x}\right)−\mathrm{1}\right)=\mathrm{0} \\ $$$${cos}\left({x}\right)=\mathrm{0}\Rightarrow\:{x}=\mathrm{2}{k}\pi\:\:;\:{k}\in\mathbb{Z} \\ $$$${cos}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\left\{_{{x}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{3}}} ^{{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{3}}} \:{k}\in\mathbb{Z}\right. \\ $$$$ \\ $$ | ||