Помогите, пожалуйста 1) 2cos^2 a = 1 + -/2 – cos 2x 2) cos^2 6x + sin^2 3x – 1 = 0

2) sin^2 3x = 1 – cos^2 6x sin^2 6x – sin^2 3x = 0 4sin^2 3x * cos^2 3x – sin^2 3x = 0 sin^2 3x*(4cos^2 3x - 1) = 0 sin^2 3x = 0    cos^2 3x =1/4 По формулам понижения степени: (1 - cos 6x)/2 = 0              (1 + cos 6x)/2 = ¼ co6x = 1                             cos6x = - ½ x = pi/3 * n                        6x = ±(2pi/3) +2pi n                                               x = ±pi/4 + pi/3 *n