Question

Тригонометрические уравнения!!! ​

Answer 1

$\sqrt{3}sin(\frac{\pi x}{6}) + cos(\frac{\pi x}{6}) - 2 = 0\\2(sin(\frac{\pi x}{6} + \frac{\pi}{6})) = 2\\sin(\frac{\pi}{6}(x+1)) = 1\\\frac{\pi}{6}(x+1) = \frac{\pi}{2}+2\pi n\\x+1 = 3 +12n\\x = 2 + 12n\\n = -1 => x = -10\\Answer: -10$

Answer 2

$\sqrt{3}Sin\frac{\pi x}{6} +Cos\frac{\pi x}{6}-2=0|:2 \\\\\frac{\sqrt{3} }{2}Sin\frac{\pi x}{6}+\frac{1}{2}Cos\frac{\pi x}{6}=1\\\\Cos\frac{\pi }{6}Sin\frac{\pi x}{6}+Sin\frac{\pi }{6}Cos\frac{\pi x}{6}=1\\\\Sin(\frac{\pi x}{6}+\frac{\pi }{6})=1\\\\\frac{\pi x}{6}+\frac{\pi }{6}=\frac{\pi }{2}+2\pi n,n\in Z$

$\frac{\pi x}{6}=\frac{\pi }{2}-\frac{\pi }{6}+2\pi n,n\in Z\\\\\frac{\pi x}{6} =\frac{\pi }{3}+2\pi n,n\in Z\\\\x=2+12n,n\in Z\\\\n=-1\Rightarrow x=-10\\\\Otvet:\boxed{x=-10}$