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Question Number 10415 by okhema francis last updated on 08/Feb/17

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

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

Answered by bansal22luvi@gmail.com last updated on 08/Feb/17

2(((cos^2 θ)/(sin^2 θ)))+cosθ=0 2cos^2 θ+cosθsin^2 θ=0 cosθ(2cosθ+sin^2 θ)=0 cosθ=0 2cosθ+sin^2 θ=0 2cosθ+1−cos^2 θ=0 cos^2 θ−2cosθ−1=0 cosθ=((2±2(√2))/2) hence cosθ=0,((1±(√2))/)

$$\mathrm{2}\left(\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\right)+\mathrm{cos}\theta=\mathrm{0} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{cos}\theta\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{cos}\theta\left(\mathrm{2cos}\theta+\mathrm{sin}^{\mathrm{2}} \theta\right)=\mathrm{0} \\ $$$$\mathrm{cos}\theta=\mathrm{0} \\ $$$$\mathrm{2cos}\theta+\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{2cos}\theta+\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\theta=\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{hence}\:\mathrm{cos}\theta=\mathrm{0},\frac{\mathrm{1}\pm\sqrt{\mathrm{2}}}{} \\ $$

Commented by okhema francis last updated on 09/Feb/17

thank you very much and let the lord be with you

$${thank}\:{you}\:{very}\:{much}\:{and}\:{let}\:{the}\:{lord}\:{be}\:{with}\:{you} \\ $$

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