Sin (5Π-x) = cos (2x+7Π)

помогите

Sin5 П*cosx-sinx*cos5 П=cos2x*cos7 П-sin2x*sin7 П

sinx=-cos2x

sinx=sin^2 (x) - cos^2 (x)

sinx=sin^2 (x) - 1+sin^2 (x)

2sin^2 (x) - sinx-1=0

sinx=y,-1<=y<=1

2y^2-y-1=0

y1=1, y2=-1/2

sinx=1

x=π/2+πn

или

sinx=-1/2

x = (-1) ^n*arcsin (-1/2) + πn = (-1) ^n * (-π/6) + πn