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Question Number 10916 by okhema last updated on 02/Mar/17

find all possible values of cosθ such that 2cot^2 θ+cosθ=0

$${find}\:{all}\:{possible}\:{values}\:{of}\:{cos}\theta\:{such} \\ $$$${that}\:\mathrm{2}{cot}^{\mathrm{2}} \theta+{cos}\theta=\mathrm{0} \\ $$

Answered by sandy_suhendra last updated on 02/Mar/17

((2cos^2 θ)/(sin^2 θ)) + cosθ =0 ((2cos^2 θ+sin^2 θcosθ)/(sin^2 θ)) = 0 ((2cos^2 θ+(1−cos^2 θ)cosθ)/(sin^2 θ)) = 0 ((2cos^2 θ+cosθ−cos^3 θ)/(sin^2 θ)) = 0 ((−cosθ(cos^2 θ−2cosθ−1))/(sin^2 θ)) = 0 (1) −cosθ=0 ⇒ cosθ=0 (2) cos^2 θ−2cosθ−1=0 cosθ = ((2±(√(4+4)))/2) = ((2±2(√2))/2) = 1±(√2) cosθ = 1+(√2) (not feasible because cosθ > 1) cosθ = 1−(√2)

$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:+\:\mathrm{cos}\theta\:=\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{cos}\theta−\mathrm{cos}^{\mathrm{3}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\frac{−\mathrm{cos}\theta\left(\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}\right)}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:−\mathrm{cos}\theta=\mathrm{0}\:\Rightarrow\:\mathrm{cos}\theta=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\mathrm{1}\pm\sqrt{\mathrm{2}}\:\:\: \\ $$$$\:\:\:\mathrm{cos}\theta\:=\:\mathrm{1}+\sqrt{\mathrm{2}}\:\left(\mathrm{not}\:\mathrm{feasible}\:\mathrm{because}\:\mathrm{cos}\theta\:>\:\mathrm{1}\right)\:\:\:\:\: \\ $$$$\:\:\:\mathrm{cos}\theta\:=\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$

Commented by okhema last updated on 03/Mar/17

thats what i got but not in surd form

$${thats}\:{what}\:{i}\:{got}\:{but}\:{not}\:{in}\:{surd}\:{form} \\ $$

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