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

f i n d a l l p o s s i b l e v a l u e s o f c o s θ s u c h

t h a t 2 {c o t}^{2} θ + c o s θ = 0

Answered by sandy_suhendra last updated on 02/Mar/17

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

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

\frac{{2 \cos}^{2} θ + (1 - \cos^{2} θ) \cos θ}{\sin^{2} θ} = 0

\frac{{2 \cos}^{2} θ + \cos θ - \cos^{3} θ}{\sin^{2} θ} = 0

\frac{- \cos θ (\cos^{2} θ - 2 \cos θ - 1)}{\sin^{2} θ} = 0

(1) - \cos θ = 0 \Rightarrow \cos θ = 0

(2) \cos^{2} θ - 2 \cos θ - 1 = 0

\cos θ = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2}

\cos θ = 1 + \sqrt{2} (not feasible because \cos θ > 1)

\cos θ = 1 - \sqrt{2}

Commented by okhema last updated on 03/Mar/17

t h a t s w h a t i g o t b u t n o t i n s u r d f o r m

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