$x^3 + 2x^2 + 5x + 2\cos x = 0$ 
$x^3 + 2x^2 + 5x + 2\cos x = 0$

How do I find the number of solutions of  this equation (in $[0, 2\pi]$) without a graph?
Attempt: 
The equation simplifies to $x(x^2 + 2x + 5)=- 2\cos x $
Minima of the quadratic occurs at $x= -1$ and it's value is $4$
Minima of $-2\cos x$ is $-2$

$$y_1=-x^3-5x$$ and $$y_2=2x^2+2\cos x$$ 
for $x\ge0,  y_1<0, y_2>0$
for $-1<x<0$ both function decrease and have only one intersection
for $x\le-1$ there are no intersection $$-x^3-5x<2x^2+2\cos x$$
