Solving $x^3-7x^2+14x-8-\frac12\sin x=0$ With the given problem:
$$x^3-7x^2+14x-8-\frac12\sin x=0$$
I factorized the cubic part:
$$x^3-7x^2+14x-8=0$$
where we test the following solutions of the fraction of the last coefficient divided by the first coefficient and its even multiplicates:
$$\pm\frac{8}{1}, \pm\frac{8}{2}, \pm\frac{8}{4}, \pm\frac{8}{6}, \pm\frac{8}{8}$$
Here, we see that positive value of the second, the third and the fifth fractions are solutions to the problem in eqn. 19. We therefore rewrite the original function to
\begin{equation}
    (x-4)(x-2)(x-1)=\frac{1}{2}\sin x
\end{equation}
But now, the situation looks only nicer, but there is no chance of solving this with any regular means. I tried to change the trigonometric part into a Taylor series, but it gets messy and I get
\begin{equation}
    (x-4)(x-2)(x-1)=\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n x^{1 + 2 n}}{(1 + 2 n)!}
\end{equation}
Any ideas what to do here?
I take a guess that the domain of solutions is 0-5, and reduce the nasty part on the right to something even nastier:
\begin{equation}
\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n x^{1 + 2 n}}{(1 + 2 n)!}=-\frac{x^{11}}{39916800}-\frac{x^9}{362880}-\frac{x^7}{5040}-\frac{x^5}{120}-\frac{x^3}{6}+x
\end{equation}
which gives on the domain   $x\in$(0,5)
\begin{equation}
    2(x-4)(x-2)(x-1)=-\frac{x^{11}}{39916800}-\frac{x^9}{362880}-\frac{x^7}{5040}-\frac{x^5}{120}-\frac{x^3}{6}+x
\end{equation}
This leads nowhere also.
 A: Here is an approximation to the zero points.

*

*We put $f_1=x^3−7x^2+14x−8$ and $f_2=\frac{1}{2}\sin x$, so that $f_1(x)=f(x)-f_2(x)$.

We then calculate the values:
$f(0)=-8\leqslant0$, $f(1.5) = 0.1263 > 0$ , $f(3) = −2.0706 < 0$, $f(5) = 12.4795 > 0$
By the mean-value theorem of continuous functions we know that $f(x) = x^3 −7x^2 + 14x −8 −0.5 \sin x = f_1(x)−f_2(x)$ has zeros in the intervals (a,b)=(0,1.5),(1.5,3) and (3,5) and no zeros if $x > 5$. See plot



*We have that $f_1(x) = x^3 −7x^2 + 14x −8 = (x −1)(x2 −6x + 8) = (x −1)(x −2)(x −4)$, that is, $f_1(x) \longrightarrow-\infty,\ x\longrightarrow-\infty,\ f_1(x)\longrightarrow-\infty,\ x\longrightarrow-\infty$.
Then, $\underset{max}{0\leqslant x\leqslant2}$ and
$f_1(x) = f_1(1.5) = 0.625 > 0.5 \sin(1.5)$ for $0.5 \sin(1.5) < 0.5$ and $\underset{2<x<4}{min}$ $f_1(x) = f_1(3) = −2 < 0.5 \sin(3)$ for $0.5 \sin(1.5) > 0 (3 < \pi)$.

Note that  $f_1(x) = x^3 −7x^2 + 14x −8$  has three real zeros, x= 1, 2
and 4, and then it can be assumed that its two local extrema are precisely between the zeros 1.5 and 3.
These properties of $f_1(x) = x^3 −7x^2 + 14x −8$ and $f_2(x) = 0.5 \sin x$ show again that the function
$f(x) = x^3 −7x^2 + 14x −8 −0.5 \sin x = f_1(x) −f_2(x)$ has a zero on each interval (a,b)= (0,1.5),
(1.5,3) and (3,5) and no zeros if $ x < 0$ and $x > 5$.
