Срочно надо решить 7cos2x+3sin²2x=3

7cos2x + 3sin^2(2x) = 3 7cos2x + 3sin^2(2x) - 3 = 0 7 cos(2x) - 3cos^(2x) = 0 cos(2x) * (3cos(2x-7)) = 0 1)  cos2x = 0 2x = π/2 + πn, n∈Z x= π/4 + (πn)/2, n∈Z 2)  3cos(2x) 7 cos(2x) = 7/3 2x = c0s^(-1)(7/3) + πk, k∈Z x2 = 1/2cos^(-1) (7/3)  + πk, k∈Z 2x = 2πm - cos^(-1)(7/3) + πm, m∈Z x3 = πm - (1/2)*cos^(-1)(7/3), m∈Z