Question

Решить уравнения: 1) 2)

Answer 1

1)
$2\cos(\frac{x}{2}-\frac{\pi}{6}) = \sqrt{3}\\ \cos(\frac{x}{2}-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\\\\ \frac{x}{2}-\frac{\pi}{6} = \frac{\pi}{6} + 2\pi n\\\\ \frac{x}{2} = \frac{\pi}{3} + 2\pi n\\\\ x = \frac{2\pi}{3} + 4\pi n\\\\\\ \frac{x}{2}-\frac{\pi}{6} = -\frac{\pi}{6} + 2\pi n\\\\ \frac{x}{2} = 2\pi n\\ x = 4\pi n$

2)
$\sin^2x - 4\sin x \cos x + 3\cos^2x=0\\ \tan^2x - 4\tan x + 3=0\\\\ \tan x = 1\\ x = \frac{\pi}{4} + \pi n\\\\ \tan x = 3\\ x = \arctan3 + \pi n$