Question Number 10415 by okhema francis last updated on 08/Feb/17
$${find}\:{all}\:{the}\:{possible}\:{values}\:{of}\:\mathrm{cos}\:\theta\:{such}\:{that}\:\mathrm{2cot}\:^{\mathrm{2}} \theta\:+\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$ \\ $$
Answered by bansal22luvi@gmail.com last updated on 08/Feb/17
$$\mathrm{2}\left(\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\right)+\mathrm{cos}\theta=\mathrm{0} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{cos}\theta\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{cos}\theta\left(\mathrm{2cos}\theta+\mathrm{sin}^{\mathrm{2}} \theta\right)=\mathrm{0} \\ $$$$\mathrm{cos}\theta=\mathrm{0} \\ $$$$\mathrm{2cos}\theta+\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{2cos}\theta+\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\theta=\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{hence}\:\mathrm{cos}\theta=\mathrm{0},\frac{\mathrm{1}\pm\sqrt{\mathrm{2}}}{} \\ $$
Commented by okhema francis last updated on 09/Feb/17
$${thank}\:{you}\:{very}\:{much}\:{and}\:{let}\:{the}\:{lord}\:{be}\:{with}\:{you} \\ $$