1+cos4x=sin3x-sinx
1+2cos²2x-1=2sinx*cos2x
2cos²2x-2sinx*cos2x=0
2cos2x(cos2x-sinx)=0
cos2x=0 cos2x-sinx=0
2x=π/2+πn 1-2sin²x-sinx=0
x=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=-1
x=(-1)^k*π/6+πk x=-π/2+2πn