Hint
The problem is better conditioned if, instead, you look for the roots of $$h(x)=\log(x)\cos(x)-\sin(x)=0$$ and the convergence of Newton method is much faster.
For example, using $x_0=4$, the iterates are : $4.09701$, $4.09546$ which is the solution for six significant digits. Using $x_0=7$, the iterates are : $7.42088$, $7.39041$, $7.39037$.
Using this transform, you will find easily the $k^{th}$ root starting at $x_0=(2k+1)\frac \pi 2$. The first iterate will just be $$x_1=\pi \left(k+\frac{1}{2}\right)-\frac{1}{\log \left(\pi
\left(k+\frac{1}{2}\right)\right)}$$
Edit
What is interesting is to look at the value of the solution $x_k$ as a function of $k$. A totally empirical model leads to the following approximation $$x_k=(2k+1)\frac \pi 2-\Big(0.0962696 +\frac{0.489665}{k^{0.409352}}\Big)$$ For example, the estimate of the $10^{th}$ root is $32.6997$ while the solution is $32.7075$; the estimate of the $50^{th}$ root is $158.455$ while the solution is $158.456$.
Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.12645,7.38901,10.5870,13.7633,16.9291$.
Edit
There is another way to generate nice approximate values of the $k^{th}$ solution. The idea is to build for function $h(x)$ its simplest Pade approximant around $\theta_k=(2k+1)\frac \pi 2$ and to compute the value of $x$ which cancels the numerator. This leads to the simple approximation $$x_k\approx\theta_k
\left(1-\frac{2 \log (\theta_k)}{\theta_k \left(2 \log ^2(\theta_
k)+1\right)-2}\right)$$ Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.13630,7.40792,10.6062,13.7814,16.9459$. The $10^{th}$ root is estimated as $32.7113$ and the $50^{th}$ root is estimated as $158.457$.
We could still improve using a Pade[1,2] approximant which would lead to $$x_k\approx \theta_k\Big(1 -\frac{3 \left(\theta_k +2 \theta_k \log ^2(\theta_k )-2\right)}{\theta_k
\log (\theta_k ) \left(5 \theta_k +6 \theta_k \log ^2(\theta_k )-12\right)-3}\Big)$$ Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.09189,7.38912,10.5942,13.7725,16.9387$. The $10^{th}$ root is estimated as $32.7074$ and the $50^{th}$ root is estimated as $158.455$.
The branches of Lambert W are the local inverses of the Elementary function $f$ with $f(z)=ze^z$, $z \in \mathbb{C}$.
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. It is also proved in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.
Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia University Press, New York, 1948
The non-elementarity of LambertW was already proved by Liouville in
Liouville, J.: Mémoire sur la classification des transcendantes et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients. Journal de mathématiques pures et appliquées 2 (1837) 56–105, 3 (1838) 5233–547
It is also proved in
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22
and in
Bronstein, M.; Corless, R. M.; Davenport, J. H., Jeffrey, D. J.: Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions 19 (2008) (10) 709-712.
Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions.
Best Answer
If you are using WolframAlpha, you will obtain the solution:
$$x=W(-5e^{-5})+5$$
Since $0>-5e^{-5} > -e^{-1}$, the Lambert-W function has two values. On WolframAlpha you can input:
for the principal branch, where $W \ge -1$. This equates to $4.96\dots$;
for the lower branch, where $W \le -1$. This equates to $0$ exactly.