Question

Тригонометричні рівняння
$\left \{ {{x-y=\frac{5\pi }{3} } \atop {sin x=2 sin y}} \right.$

Answer 1

Ответ:

$\left \{ {{x-y=\frac{5\pi }{3} } \atop {sin x=2 sin y}} \right. \\ \\\left \{ {{x=y + \frac{5\pi }{3} } \atop {sin x=2 sin y}} \right. \\ \\ \sin( \frac{5\pi}{3} + y ) = 2 \sin(y) \\ \sin( \frac{5\pi}{3} ) \cos(y) + \cos( \frac{5\pi}{3} ) \sin(y) - 2 \sin(y) = 0 \\ - \frac{ \sqrt{3} }{2} \cos(y) + \frac{ 1}{2} \sin(y) - 2\sin(y) = 0 \\ - \frac{ \sqrt{3} }{2} \cos(y) - \frac{3}{2} \sin(y) = 0 \\ - \sqrt{3} ( \frac{1}{2} \cos(y) + \frac{ \sqrt{3} }{2} \sin(y) ) = 0 \\ \sin( \frac{\pi}{6} ) \cos(y) + \cos( \frac{\pi}{6} ) \sin(y) = 0 \\ \sin(y + \frac{\pi}{6} ) = 0 \\ y + \frac{\pi}{6} = \pi \: n \\ y = - \frac{\pi}{6} + \pi \: n \\ \\ x = - \frac{\pi}{6} + \frac{5\pi}{3} + \pi \: n = 3\pi + \pi \: n \\ x = \pi \: n \\ \\ \left \{ {{y = - \frac{\pi }{6} + \pi \: n} \atop {x = \pi \: n }} \right.$

n принадлежит Z