no. of real roots of the equation $4x^5+5x^4-10x^2-20x+40=0$ 
$(1)$ Find Total no. of real solution of the equation $xe^{\sin x} = \cos x\;,$ where $\displaystyle x\in \left(0,\frac{\pi}{2}\right)$
$(2)$ The no. of real roots of the equation $4x^5+5x^4-10x^2-20x+40=0$

$\bf{My\; Try::}$ For $(1)$ st part::
Let $f(x) = xe^{\sin x}-\cos x\;,$ Then $\displaystyle f'(x) = xe^{\sin x}\cdot \cos x+e^{\sin x}+\sin x>0\;\forall x\in \left(0,\frac{\pi}{2}\right)$
So $f(x)$ is Strictly Increasing function.
And when $x\rightarrow -\infty\;,$ Then $f(x)\rightarrow -\infty$ and when $x\rightarrow +\infty\;,$ Then $f(x)\rightarrow +\infty$.
So $f(x)$ intersect the $\bf{X-}$ axis at exactly at one point.
So only one real roots of the equation.
But I did not understand how can i found that the roots lie between $\displaystyle \left(0,\frac{\pi}{2}\right)$.
Help me, Thanks
$\bf{My\; Try::}$ For $(2)$ st part::
Let $f(x)=4x^5+5x^4-10x^2-20x+40\;,$ Then $f'(x)=20x^4+20x^3-20x-20$
Now I did not understand how can i solve after that
Help me, Thanks
 A: For (2):
Your derivative is 20(x^4 +x^3 -x -1). Grouping the first two bracketed terms as one group and the last two terms as the second group, we see that x^4 +x^3 -x -1 = x^3*(x+1) -1(x+1) which is equal to (x^3-1)(x+1). This is equal to (x+1)(x-1)(x^2+x+1). Note that the term (x^2 +x+1) is always positive (I leave you to prove that fact), hence we conclude that the derivative of f(x) is negative if and only if -1< x < 1. Additionally, we note that the derivative of f is zero if and only if x is +1 or -1.
Now f(1)=19, which is clearly positive, and as the derivative of f(x) is positive for all x bigger than 1, we can be certain that f(x) has no roots in the interval [1,infinity). 
Now f(-2) is equal to -8, and f(-1) is equal to 51. So by the Intermediate Value Theorem f must have at least one root in the interval (-2, -1). Let the smallest of these roots be denoted by a. Now assume for contradiction that f has two real roots. We know one of those two roots is a, where a is in the interval (-2,-1). By Rolle's theorem, there must be a critical point of f which lies (strictly) between a and the second root. 
Now the last part of my answer is not perfectly rigorous: given that f(1) is positive and that there are only two critical points (at x=-1 and x=1), we see this is geometrically impossible, by tracing out a sketch of the curve f(x).      
A: For the second question, your derivative is incorrect. All the coefficients are $\pm20$. You can factor the derivative and identify critical points, and proceed from there considering how the function behaves between two critical points.
For the first question, one side of the equation is increasing and starts low at $x=0$, and the other is decreasing and starts high. By the time you reach $x=\pi/2$, they have swapped positions. So how many solutions can there be along the way in $(0,\pi/2)$?
A: For (2).
First, plot the polynomail to get an idea of what you are doing. Google does the job. You will see that the polynomial has only one root near $-2$. Now you must prove it.
The derivative is
$$f'(x)=20(x^4+x^3-x-1)=20(x^3-1)(x+1)$$
It has two real roots at $-1$ and $1$ as suggests the graphic of $f$. Since none of them are multiple roots, the derivative change its sign at them, so $f$ is increasing in $(-\infty,-1]$ and in $[1,\infty)$, and it is decreasing in $[-1,1]$.
Therefore, $f$ has a minimum at $1$: $f(1)=19>0$. So $f$ has only one root.
A: It is possible to bound the real roots of that poly and those bounds are [-11,11] so that makes it easier to graph it.
There is one real root at approximately -1.95
