You will have to sketch a graph to find the number of solutions.
You don't need to get it exactly detailed and in the right scale, so you can do it on rough paper pretty easily without a calculator.
But how?
You identified the first part correctly, separating the two different functions from each other:
$$x^3 + 2x^2 + 5x=- 2\cos x $$
Now, sketching the RHS on rough paper is very easy. We'll focus on the polynomial in the LHS.
Turning point of the cubic occurs at $3x^2+4x+5=0$ which has zero solutions ($\because D<0$), hence, there is no turning point i.e. the function $f(x)$ is monotonic. The coefficient of $x^3$ is $>1$, which implies the cubic starts from $-\infty$ when $x\rightarrow-\infty$, crosses the y-axis at $(0,5)$ and goes till $+\infty$ when $x\rightarrow+\infty$.
Now, $f(-1)=-1+2-5=-4<-2$ and $f(x)$ $\forall x<-1$ is less than $-2$. Hence, notice $f(x)$ has intersected $y=-\cos x$ in interval (0,1) .
Knowing the nature of LHS and RHS, we can easily comment that there will not be any more points of intersection.
Problem solved.