Найдите значения выражения √(48)-√(192)sin^2(19π/12)Решите уравнение sin(πх/4)=-1

Решите задачу:

$\sqrt{48}-\sqrt{192}*sin^2(\frac{19\pi}{12})=\sqrt{48}-\sqrt{4*48}*sin^2(\frac{19\pi}{12})=$

$\sqrt{48}(1-2*sin^2(\frac{19\pi}{12}))=\sqrt{48}*cos(\frac{19\pi}6-2\pi)=\sqrt{48}*cos(\frac{19\pi}6-\frac{12\pi}6)$

$=\sqrt{48}*cos(\frac{7\pi}6)=\sqrt{48}*(-cos(\pi-\frac{7\pi}6))=-\sqrt{48}*cos(\frac{{\pi}}6)=$

$=-\sqrt{48}*\frac{\sqrt3}2=-6$

$sin(\frac{\pi x}4)=-1$

$\frac{\pi x}4=-\frac{\pi}2+2\pi n; n \in Z$

$\frac{ x}4=-\frac{1}2+2 n; n \in Z$

$x=-2+8 n; n \in Z$