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

[latex]sqrt{48}-sqrt{192}*sin^2(frac{19pi}{12})=sqrt{48}-sqrt{4*48}*sin^2(frac{19pi}{12})=[/latex][latex]sqrt{48}(1-2*sin^2(frac{19pi}{12}))=sqrt{48}*cos(frac{19pi}6-2pi)=sqrt{48}*cos(frac{19pi}6-frac{12pi}6)[/latex][latex]=sqrt{48}*cos(frac{7pi}6)=sqrt{48}*(-cos(pi-frac{7pi}6))=-sqrt{48}*cos(frac{{pi}}6)=[/latex][latex]=-sqrt{48}*frac{sqrt3}2=-6[/latex][latex]sin(frac{pi x}4)=-1[/latex][latex]frac{pi x}4=-frac{pi}2+2pi n; n in Z[/latex][latex]frac{ x}4=-frac{1}2+2 n; n in Z[/latex][latex]x=-2+8 n; n in Z[/latex]