sin(5Π-x)=cos(2x+7Π)
помогите

sin5П*cosx-sinx*cos5П=cos2x*cos7П-sin2x*sin7Пsinx=-cos2xsinx=sin^2(x)-cos^2(x)sinx=sin^2(x)-1+sin^2(x)2sin^2(x)-sinx-1=0sinx=y,-1<=y<=12y^2-y-1=0y1=1,y2=-1/2sinx=1x=π/2+πnилиsinx=-1/2x=(-1)^n*arcsin(-1/2)+πn=(-1)^n*(-π/6)+πn