If-x-2-tan-cot-x-1-0-has-two-real-solutions-2-3-2-3-find-sin-cos-

Question Number 135791 by benjo_mathlover last updated on 16/Mar/21

I f x^{2} + (\tan θ + \cot θ) x + 1 = 0

h a s t w o r e a l s o l u t i o n s

{2 - \sqrt{3}, 2 + \sqrt{3}}, f i n d {\begin{cases} \sin θ \\ \cos θ \end{cases} .

Answered by EDWIN88 last updated on 17/Mar/21

By {Vieta}^{'} s \Rightarrow x_{1} + x_{2} = - \frac{b}{a}

\Rightarrow \tan θ + \cot θ = - 4, \tan θ + \frac{1}{\tan θ} = - 4

\tan^{2} θ + 4 \tan θ + 1 = 0

{(\tan θ + 2)}^{2} - 3 = 0 \Rightarrow \tan θ = - 2 \pm \sqrt{3}

or \tan^{2} θ = 7 \pm 4 \sqrt{3} \to {\begin{cases} \sec^{2} θ = 8 \pm 4 \sqrt{3} \\ \cos^{2} θ = \frac{1}{8 \pm 4 \sqrt{3}} \end{cases}

{\begin{cases} \cos θ = \pm \sqrt{\frac{1}{{(\sqrt{6} \pm \sqrt{2})}^{2}}} = \pm \frac{1}{\sqrt{6} \pm \sqrt{2}} \\ \sin^{2} θ = 1 - \cos^{2} θ = 1 - \frac{1}{8 \pm 4 \sqrt{3}} = \frac{7 \pm 4 \sqrt{3}}{8 \pm 4 \sqrt{3}} \end{cases}

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