(1+cos2x)sinx=cos^2x
sin3x=-2cosx
sin3x-2sinxcos2x=0

(1+2cos^2(x) -1)sinx=cos^2(x) cos^2(x)*(2sinx-1)=0 Два случая: 1) cos^2(x)=0 <=> cosx=0 <=> x=pi/2+pi*k 2) 2sinx-1=0 <=> sinx=1/2 <=> x=(-1)^n*pi/6+pi*n sin3x-2sinxcos2x=0 sin2x*cosx+sinx*cos2x-2sinxcos2x=0 sin2xcosx-sinxcos2x=0 sin(2x-x)=0 sinx=0 <=> x=pi*m