Корень из 3 sin2x-2 cos^2 x=2 корень из (2+2cos2x)

√3*sin(2x) - 2cos^2(x) = 2√(2+2cos(2x))√3*2sinx*cosx - 2cos^2(x) = 2√(2+2(2cos^2(x) - 1))2cosx*(√3*sinx - cosx) = 2√(2+4cos^2(x) - 2) = 2√(4cos^2(x)) = 4*|cosx|Разбиваем на две системы, раскрывая модуль:1) cosx ≥ 02cosx*(√3*sinx - cosx) = 4cosx2cosx*(√3*sinx - cosx - 2) = 0cosx = 0, x = π/2 + πk, k∈Zsin(2*x/2) = 2*sin(x/2)*cos(x/2)cos(2*x/2) = cos^2(x/2) - sin^2(x/2)2 = 2cos^2(x/2) + 2sin^2(x/2)√3*2*sin(x/2)*cos(x/2) - cos^2(x/2) + sin^2(x/2) - 2cos^2(x/2) - 2sin^2(x/2) = 0-3cos^2(x/2) - sin^2(x/2) + 2√3*sin(x/2)*cos(x/2) = 0 - разделим обе части на cos^2(x/2)-3 - tg^2(x/2) + 2√3*tg(x/2) = 0tg^2(x/2) - 2√3*tg(x/2) + 3 = 0, tg(x/2) = tt^2 - 2√3*t + 3 = 0, D=4*3 - 4*3 = 0t = √3, tg(x/2) = √3, (x/2) = π/3 + πk, x = 2π/3 + 2πk, k∈Z2) cosx < 02cosx*(√3*sinx - cosx + 2) = 0cosx = 0 - не учитываем, т.к. неравенство строгое.(√3*sinx - cosx + 2 = 0) - преобразуем аналогично первому пункту, получим:√3*2*sin(x/2)*cos(x/2) - cos^2(x/2) + sin^2(x/2) + 2cos^2(x/2) + 2sin^2(x/2) = 0cos^2(x/2) + 3sin^2(x/2) + 2√3*sin(x/2)*cos(x/2) = 01 + 3tg^2(x/2) + 2√3*tg(x/2) = 03t^2 + 2√3*t + 1 = 0, D=4*3 - 4*3 = 0t = -2√3/6 = -√3/3tg(x/2) = -√3/3, (x/2) = -π/6 + πk, x = -π/3 + 2πk, k∈ZОбъединяем три решения, получаем: x = π/2 + πk, x = 2π/3 + πk, k∈Z