sinx/2*sin3x/2=1/2
1/2[cos(x/2 - 3x/2) - cos(x/2 + 3x/2)] = 1/2
cos(x) - cos(2x) = 1
применяем формулу: (cos2x = 2cos²x - 1)
2cos²x - cosx = 0
cosx(2cosx - 1) = 0
1) cosx = 0
x₁ = π/2 + πk, k∈Z
2) 2cosx - 1 = 0
cosx = 1/2
x = (+ -)arccos(1/2) + 2πn, n∈Z
x₂ = (+ -)(π/3) + 2πn, n∈Z