Solve the equation on the interval [0,

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