Question Number 10916 by okhema last updated on 02/Mar/17
$${find}\:{all}\:{possible}\:{values}\:{of}\:{cos}\theta\:{such} \\ $$$${that}\:\mathrm{2}{cot}^{\mathrm{2}} \theta+{cos}\theta=\mathrm{0} \\ $$
Answered by sandy_suhendra last updated on 02/Mar/17
$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:+\:\mathrm{cos}\theta\:=\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{cos}\theta−\mathrm{cos}^{\mathrm{3}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{−\mathrm{cos}\theta\left(\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}\right)}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:−\mathrm{cos}\theta=\mathrm{0}\:\Rightarrow\:\mathrm{cos}\theta=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\mathrm{1}\pm\sqrt{\mathrm{2}}\:\:\: \\ $$$$\:\:\:\mathrm{cos}\theta\:=\:\mathrm{1}+\sqrt{\mathrm{2}}\:\left(\mathrm{not}\:\mathrm{feasible}\:\mathrm{because}\:\mathrm{cos}\theta\:>\:\mathrm{1}\right)\:\:\:\:\: \\ $$$$\:\:\:\mathrm{cos}\theta\:=\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$
Commented by okhema last updated on 03/Mar/17
$${thats}\:{what}\:{i}\:{got}\:{but}\:{not}\:{in}\:{surd}\:{form} \\ $$