Помогите плиз(((cos π(x-3)/6=-^3/2

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

$cos\frac{\pi x-3\pi }{6} =-\frac{\sqrt{3} }{2}$

$cos2\pi \frac{\pi x-3\pi }{6} =-\frac{\sqrt{3} }{2}$

$\frac{\pi x -3\pi }{6}=arccos-\frac{\sqrt{3} }{2}$

$2\pi- \frac{\pi x -3\pi }{6}=arccos-\frac{\sqrt{3} }{2}$

$\frac{\pi x-3\pi }{6}=\frac{5\pi }{6}$ +2kπ, k∈Z

$2\pi -\frac{\pi x-3\pi }{6}=\frac{5\pi }{6}$ +2kπ, k∈Z

$x=8+12k$ , k∈Z

$x=10-12k$ , k∈Z

ОТВЕТ:

$x=\left \{ {{8+12k} \atop {10+12k}} \right.$ , k∈Z