1) cos x√2

2 = 2

2) sin x sin 2x+cos3x=0

3) 2cos x+cos²x=x-sin²x

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