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

f i n d a l l t h e p o s s i b l e v a l u e s o f \cos θ s u c h t h a t 2 \cot^{2} θ + \cos θ = 0

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

2 (\frac{\cos^{2} θ}{\sin^{2} θ}) + \cos θ = 0

{2 \cos}^{2} θ + \cos θ \sin^{2} θ = 0

\cos θ (2 \cos θ + \sin^{2} θ) = 0

\cos θ = 0

2 \cos θ + \sin^{2} θ = 0

2 \cos θ + 1 - \cos^{2} θ = 0

\cos^{2} θ - 2 \cos θ - 1 = 0

\cos θ = \frac{2 \pm 2 \sqrt{2}}{2}

hence \cos θ = 0, \frac{1 \pm \sqrt{2}}{}

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

t h a n k y o u v e r y m u c h a n d l e t t h e l o r d b e w i t h y o u