Question

(1+cos2x)sinx=cos^2xsin3x=-2cosxsin3x-2sinxcos2x=0

Answer 1

(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