Решите уравнение
(Корень из 3*sin2x+cos2x)/2=1

(√3/2)*sin2x + (1/2)*cos2x = 1sin(π/3)sin2x + cos(π/3)cos2x = 1cos(π/3 - 2x) = 1cos(2x - π/3) = 12x - π/2 = 2πk, k∈Z2x = π/2 + 2πk, k∈Zx = π/4 + πk, k∈Z