Question

cos2x + cosx + sinx=0
помогите пожалуйста

Answer 1

2cos^2 x - 1 + cos x + sin x = 0
2(2cos^2 (x/2) - 1)^2 - 1 + (2cos^2 (x/2) - 1) + 2sin(x/2)*cos(x/2) = 0
2(4cos^4 (x/2)-4cos^2 (x/2)+1) - 1 + 2cos^2 (x/2) - 1 + 2sin(x/2)*cos(x/2) = 0
8cos^4 (x/2) - 8cos^2 (x/2) + 2 + 2cos^2 (x/2) - 2 + 2sin(x/2)*cos(x/2) = 0
8cos^4 (x/2) - 6cos^2 (x/2) + 2sin(x/2)*cos(x/2) = 0
2cos (x/2)*(4cos^3 (x/2) - 3cos (x/2) + sin (x/2)) = 0
1) cos x/2 = 0; x/2 = pi/2 + pi*k;
x1 = pi + 2pi*k

2) 4cos^3 (x/2) - 3cos (x/2) + sin (x/2) = 0
Заметим, что 4cos^3 a - 3cos a = cos 3a. Получаем:
cos (3x/2) + sin (x/2) = 0
cos (3x/2) + cos (pi/2 - x/2) = 0
Применим формулу суммы косинусов
$2cos frac{3x/2+ pi /2-x/2}{2}*cos frac{3x/2- pi /2+x/2}{2} =0$
$2cos (frac{x}{2}+ frac{ pi }{4} )*cos( x - frac{ pi }{4} )=0$

3) cos (x/2 + pi/4) = 0; x/2 + pi/4 = pi/2 + pi*n;
x2 = pi/2 + 2pi*n

4) cos (x - pi/4) = 0; x - pi/4 = pi/2 + pi*m
x3 = 3pi/4 + pi*m