Решите уравнение: [latex]sin5x- \sqrt{3} cos5x=2[/latex] (вроде путем введения доп. угла)

[latex]sin5x-\sqrt3cos5x=2\; |:2\\\\\frac{1}{2}sin5x-\frac{\sqrt3}{2}cos5x=1\\\\cos\frac{\pi}{3}\cdot sin5x-sin\frac{\pi}{3}\cdot cos5x=1\\\\sin(5x-\frac{\pi}{3})=1\\\\5x-\frac{\pi}{3}=\frac{\pi}{2}+2\pi n,\; n\in Z\\\\5x=\frac{\pi}{3}+\frac{\pi}{2}+2\pi n=\frac{5\pi}{6}+2\pi n,\; n\in Z\\\\x=\frac{\pi}{6}+\frac{2\pi n}{5},\; n\in Z[/latex]