Question

100 pontos.Me expliquem passo a passo cm fazer as contas da questao

Answer 1

$\left[\begin{array}{cc}3&6\\6&8\end{array}\right] \left[\begin{array}{c}tg \alpha\\cos \beta\end{array}\right] = \left[\begin{array}{c}0\\-2\sqrt3\end{array}\right] \\ \\ 3tg\alpha+6cos \beta=0\\ 6tg \alpha+8cos \beta=-2\sqrt3\\ \\ 6tg \alpha+12cos\beta=0\\ 6tg \alpha+8cos \beta=-2\sqrt3\\ \\ 4cos\beta=2\sqrt3\\ cos \beta=\frac{\sqrt3}{2}\\ \\ \beta=\frac{\pi}{6} \ rad\\ \\ 3tg \alpha+6.\frac{\sqrt3}{2}=0\\ 3tg \alpha=-3\sqrt3\\ \\ tg \alpha=-sqrt3\\ \\ \alpha=\frac{5\pi}{3}$

$\alpha + \beta=\frac{-\pi}{6}+\frac{\pi}{3}=\frac{-\pi+2\pi}{6}=\frac{\pi}{6}$

Answer 2

$\dbinom {3~~6} {6~~8}.\dbinom {tg \alpha} {cos \beta}=\dbinom {0} {-2 \sqrt{3} }\to\dbinom {3tg\alpha+6cos\beta} {6tg \alpha+8cos\beta}= \dbinom {0} {-2 \sqrt{3} } \\ \\ \\\begin {cases} 3tg\alpha+6cos\beta=0~~. (-2)\\ 6tg \alpha+8cos\beta =-2 \sqrt{3} \\ \end {cases} \to \begin {cases} -6tg\alpha-12cos\beta=0\\ 6tg \alpha+8cos\beta =-2 \sqrt{3} \\ \end {cases} (+) \\ \\ -4cos \beta=-2 \sqrt{3}\to cos~\beta =\dfrac{-2 \sqrt{3} }{-4}= \dfrac{ \sqrt{3} }{2}\therefore \beta =30\°=\frac{\pi }{6}$$6tg\alpha+8cos\beta=-2 \sqrt{3} \\ 6tg \alpha+8 .\dfrac{ \sqrt{3} }{2} =-2 \sqrt{3} \\ 6tg \alpha=-2 \sqrt{3} -4 \sqrt{3} \\ 6tg \alpha =-6 \sqrt{3} \\ \alpha =- \sqrt{3} =300\°=-60\°=- \dfrac{ \pi }{3} \\ \\ \alpha + \beta =- \dfrac{ \pi }{3} +\dfrac{ \pi }{6}= -\dfrac{2 \pi + \pi }{6} \therefore \alpha + \beta = -\dfrac{ \pi }{6} \\ \\ \\ .$