cos2x + cosx + sinx=0
помогите пожалуйста

2cos^2 x - 1 + cos x + sin x = 02(2cos^2 (x/2) - 1)^2 - 1 + (2cos^2 (x/2) - 1) + 2sin(x/2)*cos(x/2) = 02(4cos^4 (x/2)-4cos^2 (x/2)+1) - 1 + 2cos^2 (x/2) - 1 + 2sin(x/2)*cos(x/2) = 08cos^4 (x/2) - 8cos^2 (x/2) + 2 + 2cos^2 (x/2) - 2 + 2sin(x/2)*cos(x/2) = 08cos^4 (x/2) - 6cos^2 (x/2) + 2sin(x/2)*cos(x/2) = 02cos (x/2)*(4cos^3 (x/2) - 3cos (x/2) + sin (x/2)) = 01) cos x/2 = 0; x/2 = pi/2 + pi*k; x1 = pi + 2pi*k2) 4cos^3 (x/2) - 3cos (x/2) + sin (x/2) = 0Заметим, что 4cos^3 a - 3cos a = cos 3a. Получаем:cos (3x/2) + sin (x/2) = 0cos (3x/2) + cos (pi/2 - x/2) = 0Применим формулу суммы косинусов3) cos (x/2 + pi/4) = 0; x/2 + pi/4 = pi/2 + pi*n; x2 = pi/2 + 2pi*n4) cos (x - pi/4) = 0; x - pi/4 = pi/2 + pi*mx3 = 3pi/4 + pi*m