If x^2 +(tan θ+cot θ)x + 1 = 0 has two real solutions { 2−(√3) , 2+(√3) }, find { ((sin θ)),((cos θ)) :} .


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Question Number 135791 by benjo_mathlover last updated on 16/Mar/21
$If x^2 +(tan θ+cot θ)x + 1 = 0 has two real solutions { 2−(√3) , 2+(√3) }, find { ((sin θ)),((cos θ)) :} .$
$I f x^{2} + (\tan θ + \cot θ) x + 1 = 0$ $h a s t w o r e a l s o l u t i o n s$ ${2 - \sqrt{3}, 2 + \sqrt{3}}, f i n d {\begin{cases} \sin θ \\ \cos θ \end{cases} .$
	Answered by EDWIN88 last updated on 17/Mar/21
	$By Vieta′s ⇒ x_1 +x_2 = −(b/a) ⇒tan θ+cot θ = −4 , tan θ+(1/(tan θ))=−4 tan^2 θ+4tan θ+1 = 0 (tan θ+2)^2 −3 = 0 ⇒tan θ = −2±(√3) or tan^2 θ = 7±4(√3) → { ((sec^2 θ = 8±4(√3))),((cos^2 θ=(1/(8±4(√3))))) :} { ((cos θ = ±(√(1/(((√6) ±(√2))^2 ))) = ± (1/( (√6)±(√2))))),((sin^2 θ=1−cos^2 θ=1−(1/(8±4(√3))) = ((7±4(√3))/(8±4(√3))))) :}$
	$By {Vieta}^{'} s \Rightarrow x_{1} + x_{2} = - \frac{b}{a}$ $\Rightarrow \tan θ + \cot θ = - 4, \tan θ + \frac{1}{\tan θ} = - 4$ $\tan^{2} θ + 4 \tan θ + 1 = 0$ ${(\tan θ + 2)}^{2} - 3 = 0 \Rightarrow \tan θ = - 2 \pm \sqrt{3}$ $or \tan^{2} θ = 7 \pm 4 \sqrt{3} \to {\begin{cases} \sec^{2} θ = 8 \pm 4 \sqrt{3} \\ \cos^{2} θ = \frac{1}{8 \pm 4 \sqrt{3}} \end{cases}$ ${\begin{cases} \cos θ = \pm \sqrt{\frac{1}{{(\sqrt{6} \pm \sqrt{2})}^{2}}} = \pm \frac{1}{\sqrt{6} \pm \sqrt{2}} \\ \sin^{2} θ = 1 - \cos^{2} θ = 1 - \frac{1}{8 \pm 4 \sqrt{3}} = \frac{7 \pm 4 \sqrt{3}}{8 \pm 4 \sqrt{3}} \end{cases}$
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