1+cos4x=sin3x-sinx1+2cos²2x-1=2sinx*cos2x
2cos²2x-2sinx*cos2x=0
2cos2x(cos2x-sinx)=0cos2x=0 cos2x-sinx=02x=π/2+πn 1-2sin²x-sinx=0x=π/4+πn/2 2sin²x+sinx-1=0
Пусть sinx = t
2t²+t-1=0
D=1+8=9; √D=3
t1=(-1+3)/4=1/2
t2=(-1-3)/4=-1
Замена
sinx=1/2 sinx=-1x=(-1)^k*π/6+πk x=-π/2+2πn