Question

Найдите наибольший отрицательный корень уравнения.
Помогите пожалуйста.

Answer 1

$sin\frac{\pi x}{6}-\sqrt3\, cos\frac{\pi x}{6}=2\; \Big |:2\\\\\frac{1}{2}\cdot sin\frac{\pi x}{6}-\frac{\sqrt3}{2}\cdot cos\frac{\pi x}{6}=1\\\\sin\frac{\pi }{6}\cdot sin\frac{\pi x}{6}-cos\frac{\pi }{6}\cdot cos\frac{\pi x}{6}=1\\\\-cos(\frac{\pi }{6}+\frac{\pi x}{6})=1\\\\cos(\frac{\pi }{6}+\frac{\pi x}{6})=-1\\\\\frac{\pi }{6}+\frac{\pi x}{6}=\pi +2\pi n,\; n\in Z\\\\\frac{\pi x}{6}=\frac{5\pi }{6}+2\pi n,\; n\in Z\\\\x=5+12n\; ,\; n\in Z$

$n=0:\; \; x=5\\\\n=-1:\; \; x=5-12=-7\\\\n=-2:\; \; x=5-24=-19\\\\..............................................\\\\Otvet:\; \; x_{naibol.otr.}=-7$