3sinx^2x-2корень из 3*sinx*cosx+cos^2x=0

√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 ≥ 0 2cosx*(√3*sinx - cosx) = 4cosx 2cosx*(√3*sinx - cosx - 2) = 0 cosx = 0, x = π/2 + πk, k∈Z sin(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) = 0 tg^2(x/2) - 2√3*tg(x/2) + 3 = 0, tg(x/2) = t t^2 - 2√3*t + 3 = 0, D=4*3 - 4*3 = 0 t = √3, tg(x/2) = √3, (x/2) = π/3 + πk, x = 2π/3 + 2πk, k∈Z 2) cosx < 0 2cosx*(√3*sinx - cosx + 2) = 0 cosx = 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) = 0 cos^2(x/2) + 3sin^2(x/2) + 2√3*sin(x/2)*cos(x/2) = 0 1 + 3tg^2(x/2) + 2√3*tg(x/2) = 0 3t^2 + 2√3*t + 1 = 0, D=4*3 - 4*3 = 0 t = -2√3/6 = -√3/3 tg(x/2) = -√3/3, (x/2) = -π/6 + πk, x = -π/3 + 2πk, k∈Z Объединяем три решения, получаем: x = π/2 + πk, x = 2π/3 + πk, k∈Z (ВРОДЕБЫ ТАК)