algebra precalculus - 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

Answer

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).