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

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 $-2$


Let $$f(x)=x^3 + 2x^2 + 5x + 2\cos x$$ 

Then $$f'(x)=3x^2+4x-2\sin x+5$$  Since $$3x^2+4x+5>2\ge 2\sin x$$ for all $x\in\mathbb{R}$, we have that $$f'(x)>0$$ for all $x\in\mathbb{R}$. Thus $f$ is monotonically increasing for all real $x$.
Now as $$f(-1)=-1+2-5+1.08...=-2.91...<0,$$ $$f(0)=0+0+0+2=2>0$$ and $f$ is continuous for all values of $x$, by the Intermediate Value Theorem the equation $$f(x)=0$$ has only one solution.
P.S. If you are interested, the value of $x$ such that $f$ crosses the $x$-axis is around $-0.421$: https://www.desmos.com/calculator/2vnhsxqgz2
