Помогите в решении: sin 5x = sin x + sin 2x

Sin (5x) = sin (x) + sin (2x) sin (5x) = sin (2x + 3x) = sin (2x) * cos (3x) + cos (2x) * sin (3x) = = 2sin (x) * cos (x) * cos (3x) + cos (2x) * (3sin (x) - 4sin^3 (x)) = = sin (x) * [2cos (x) * cos (3x) + 3cos (2x) - 4cos (2x) * sin^2 (x) ] = = sin (x) * [2cos (x) * cos (3x) + 3cos (2x) - 4cos (2x) * (1 - cos (2x)) / 2] = = sin (x) * [2cos (x) * cos (3x) + 3cos (2x) - 2cos (2x) + 2cos^2 (2x) ] = = sin (x) * [2cos (x) * cos (3x) + cos (2x) + 2cos^2 (2x) ] Получаем sin (x) * [2cos (x) * cos (3x) + cos (2x) + 2cos^2 (2x)) ] = sin (x) + 2sin (x) * cos (x) = sin (x) * (1 + 2cos (x)) 1) sin x = 0, x1 = pi*k 2) 2cos (x) * cos (3x) + cos (2x) + 2cos^2 (2x) = 1 + 2cos (x) 2cos (x) * cos (3x) + cos (2x) + 2cos^2 (2x) - 2cos (x) - 1 = 0 2cos (x) * cos (3x) + cos (2x) + 2cos^2 (2x) - 1 - 2cos (x) = 0 2cos (x) * cos (3x) - 2cos (x) + cos (2x) + cos (4x) = 0 2cos (x) * (cos (3x) - 1) + 2cos (3x) * cos x = 0 2cos x * (cos (3x) - 1 + cos (3x)) = 0 cos x * (2cos (3x) - 1) = 0 3) cos x = 0, x2 = pi/2 + pi*k cos (3x) = 1/2, x3 = 1/3 * (pi/3 + 2pi*k) = pi/9 + 2pi/3*k, x4 = 1/3 * (-pi/3 + 2pi*k) = - pi/9 + 2pi/3*k Ответ: x1 = pi*k, x2 = pi/2 + pi*k, x3 = pi/9 + 2pi/3*k, x4 = - pi/9 + 2pi/3*k