Question

Cos^2x+0,1cosx=0
Помогите решать уравнение

Answer 1

${cos}^{2} x + 0.1cosx = 0 \\ cosx(cosx + 0.1) = 0 \\ \\ cosx = 0 \\ x = \frac{\pi}{2} + \pi n \\ \\ cosx + 0.1 = 0 \\ cosx = - 0.1 \\ x = + - arccos( - 0.1) + 2\pi n \\ x = + - (\pi - arccos0.1) + 2\pi n$
Ответ: pi/2 + pi*n, n € Z; +- (pi - arccos0,1) + 2pi*n, n € Z.