Question

2(sin^3x+cos^3x)=sin2x(sinx+cos x)

Answer 1

2sin^3x-2sinx+1-sin^2x=0

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

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

sin^2x=1 sinx=1  x1=-3,5П

              sinx=-1  x2=-2,5П

 sinx=1/2           x3=-19П/6