Question

СРОЧНО!!!!! Помогите

Answer 1

$2sin(2x+\frac{\pi}{6}) - cos(x) = \sqrt{3}sin(2x) - 1\\2(sin(2x)*\frac{\sqrt{3}}{2} + cos(2x)*\frac{1}{2}) - cos(x) = \sqrt{3}sin(2x) - 1\\\sqrt{3}sin(2x) + cos(2x) - cos(x) = \sqrt{3}sin(2x) - 1\\cos(2x)-cos(x) = -1\\2cos^2(x) - 1 - cos(x) = -1\\cos(x)(2cos(x)-1) = 0\\cos(x) = 0 => x = \frac{\pi}{2} + \pi n, n \in Z\\cos(x) = \frac{1}{2} => x = (+-)\frac{\pi}{3} + 2\pi n, n \in Z$