Помогите решить уравнения a) 1+cos2x=sin2x; b) sin3xsinx+cos3xcosx=-1.

1) sin^2x+cos^2x+cos^2x-sin^2x = 2sinxcosx

cos^2x = sinxcosx

cosx (cosx-sinx) = 0

cosx = 0 = > x = π/2 + πn

cosx = sinx = > 1 = tgx = > x = π/4 + πn

2) sin3xsinx+cos3xcosx=1

cos (3x-x) = 1

cos2x = 1

2x = 2 πn

x = πn