Решите уравнение 2sin^(2)x+7cosx+2=0

2sin^(2) x + 7cos x + 2 = 0; 2 * (1 - cos^(2) x ) + 7cos x + 2 = 0; 2 - 2cos^(2) x + 7cos x + 2 = 0; - 2cos^(2) x + 7cos x + 4 = 0; 2cos^(2) x - 7cos x - 4 = 0; cos x = t — замена; 2t^2 - 7t - 4 =0; D = (-7)^2 - 4 * (-4) * 2 = 49 + 32 = 81; корень из D = 9; t1 = (7 + 9)/(2 * 2) = 16/4 = 4; t2 = (7-9)/(2 * 2) = -2/4 = -1/2; cos x = 4 — не существует; cos x = -1/2; x = +- arccos( -1/2) + 2 * Pi * n, n Є Z; x = +- (Pi - arccos(1/2)) + 2 * Pi * n, n Є Z; x = +- (Pi - Pi/3) + 2 * Pi * n, n Є Z; x = +- (2Pi/3) + 2 * Pi * n, n Є Z. Ответ: x = +- (2Pi/3) + 2 * Pi * n, n Є Z.