Question

тригонометрия
найти
$\sin(2 \alpha )$
если
$\frac{ \sin( \alpha ) +2 \cos( \alpha ) }{2 \sin( \alpha ) - 3 \cos( \alpha ) } = - 2$
​

Answer 1

$\sin\alpha+2\cos\alpha = -4\sin\alpha+6\cos\alpha\\\\5\sin\alpha = 4\cos\alpha\\\\\tan\alpha = 4/5$

Есть известная формула

$\displaystyle \sin2\alpha = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha= \frac{2\tan\alpha}{1+\tan^2\alpha}=\frac{8/5}{1+16/25} = \frac{40}{41}$

Answer 2

$\displaystyle\bf\\\frac{Sin\alpha +2Cos\alpha }{2Sin\alpha -3Cos\alpha } =-2\\\\\\\frac{Sin\alpha +2Cos\alpha }{2Sin\alpha -3Cos\alpha } +2=0\\\\\\\frac{Sin\alpha +2Cos\alpha+4Sin\alpha-6Cos\alpha }{2Sin\alpha -3Cos\alpha } =0\\\\\\\frac{5Sin\alpha -4Cos\alpha }{2Sin\alpha -3Cos\alpha } =0\\\\\\\frac{\dfrac{5Sin\alpha }{Cos\alpha } -\dfrac{4Cos\alpha }{Cos\alpha } }{\dfrac{2Sin\alpha }{Cos\alpha } -\dfrac{3Cos\alpha }{Cos\alpha } } =0$

$\displaystyle\bf\\\frac{5tg\alpha -4}{2tg\alpha-3 } =0\\\\\\5tg\alpha -4=0 \ \ \ ; \ \ tg\alpha \neq 1,5\\\\\\5tg\alpha =4\\\\\\tg\alpha =0,8\\\\\\Sin2\alpha =\frac{2tg\alpha }{tg^{2}\alpha +1 } =\frac{2\cdot 0,8}{0,8^{2} +1} =\\\\\\=\frac{1,6}{0,64+1} =\frac{1,6}{1,64} =\frac{160}{164} =\frac{40}{41} \\\\\\Otvet \ : \ Sin2\alpha =\frac{40}{41}$