Solve the equation on the interval [0, 2_).

cos x + 2 cos x sin x = 0

cos x + 2 cos x sin x = 0, this is equivalent to cos x (1 + 2 sin x) = 0, so

we have cosx=0 or 1 + 2 sin x = 0,

cosx=0, x = arccos (o) = pi/2 + 2kpi

1 + 2 sin x = 0, sinx = - 1/2, implies x=arcsin ( - 1/2) = - pi/6 + 2kpi

the solution is x = pi/2 + 2kpi, or x = - pi/6 + 2kpi, k is a relative integer