1) cos x/2=√2/2, x/2=±arccos√2/2+2πn,n∈Z, x=±2π/4+4πn, x=±π/2+4πn
2)sinx sin2x +cos3x=0
Формула: sinα * sinβ=1/2 [cos(α-β)-cos(α+β)]
1/2 *[cos(-x) - cos3x ] + cos3x=0, Учтем чётность косинуса cos(-x)=cosx
1/2* cosx+1/2* cos3x=0
1/2* (cosx+cos3x)=0
1/2* 2 * cos2x * cosx=0
cos2x=0, 2x=π/2+πn, x=π/4+πn/2,n∈Z
cosx=0, x=π/2+πk, k∈Z
3) 2cosx+cos²x=-sin²x
2cosx+(cos²x+sin²x)=0, 2cosx+1=0,
cosx=-1/2, x=±arccos(-1/2)+2πn, x=±(π-π/3)+2πn, x=±2π/3+2πn, n∈Z
