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 ≥ 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