Помогите решить 3 примера!cos(pi/9-4x)=1sinx/7=0sinx-√3cosx=1

Решите задачу:

$1.\\cos(\frac{\pi}{9}-4x)=1\\4x-\frac{\pi}{9}=2\pi n\\4x=\frac{\pi}{9}+2\pi n\\x=\frac{\pi}{36}+\frac{\pi n}{2}, \; n\in Z.$

$2.\\sin\frac{x}{7}=0\\\frac{x}{7}=\pi n\\x=7 \pi n, \; n\in Z;\\\\ 3.\\sinx-\sqrt3cosx=1| \; :2\\\frac{1}{2}sinx-\frac{\sqrt3}{2}cosx=\frac{1}{2}\\\frac{1}{2}=cos\frac{\pi}{3}, \quad \frac{\sqrt3}{2}=sin\frac{\pi}{3};\\sinxcos\frac{\pi}{3}+cosxsin\frac{\pi}{3}=\frac{1}{2}\\sin(x+\frac{\pi}{3})=\frac{1}{2}\\x+\frac{\pi}{3}=(-1)^{n}arcsin\frac{1}{2}+\pi n\\x=(-1)^n\frac{\pi}{6}-\frac{\pi}{3}+\pi n, \; n\in Z;$

$x+\frac{\pi}{3}=\frac{\pi}{6}+2\pi n\\x=\frac{\pi}{2}+2\pi n, \; n\in Z;\\\\x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+2\pi n\\x+\frac{\pi}{3}=\frac{5\pi}{6}+2\pi n\\x=\frac{\pi}{2}+2\pi n, \; n\in Z.$