Solve $\frac{1}{x}\cdot \cos x + \ln x \sin x = 0$ I was working on the following function:

$$f(x) = \frac{\ln x}{\cos x}$$

I tried to find values of x where derivative will equal to zero.
After taking derivative of $f'(x)$, I got:
$$\tag 1f'(x) = \frac{\frac{1}{x} \cdot \cos x + \ln x \cdot \sin x}{\cos^2 x}$$
Setting numerator to zero, we get $$\tag 2\frac{1}{x} \cdot \cos x + \ln x \cdot \sin x = 0 $$
And now I'm stuck. How can I solve equation above (2) algebraically?

I've made some attempts at solving this myself, but, as you can probably tell, they led me nowhere. 
 A: Although no exact analytical solutions for this mixed log-trigonometric equation are available, really good analytic approximations can still be derived. 
Rewrite the equation 
$$\frac{\cos x}{x} + \ln x\sin x = 0 $$
equivalently as,
$$ \tan x =-\frac{1}{x\ln x}$$
Observe that the rhs quickly becomes small as $x$ moves out. Since $\tan(x)$ assumes small values around $k\pi$, there would be infinite number of roots, all around $k\pi$. 
To proceed, let $x=k\pi +y$ and approximate $\tan(x)$ around $k\pi$ as 
$$\tan(x)=\tan(k\pi+y) \approx y \tag{1}$$
and, similarly, approximate $-1/(x\ln x)$  around $k\pi$ as
$$-\frac{1}{x\ln x} \approx 
-\frac{1}{k\pi\ln (k\pi)} 
+\frac{\ln(k\pi)+ 1}{[k\pi \ln (k\pi)]^2}
y\tag{2} $$
As a result, $y$ can be solved from (1) and (2),
$$y_k =
-\frac{k\pi\ln (k\pi)}
{[k\pi \ln (k\pi)]^2-\ln(k\pi) - 1}$$
And, hence, the solutions to the original equation $ x_k = k\pi + y_k$,
$$ x_k = k\pi \left[1-\frac{\ln (k\pi)}
{(k\pi)^2\ln^2 (k\pi)-\ln(k\pi) - 1} \right]\tag{3}$$
with $k=1,2,3,  ... \infty$. 
For illustration, the first few roots are 
$$x_1 \approx \pi - 0.33334 = 2.80825 \space (2.80984)$$
$$x_2 \approx 2\pi - 0.08848= 6.19471 \space (6.19490)$$
$$x_3 \approx 3\pi - 0.04764 =9.37714 \space (9.37717)$$
$$...$$
$$x_n=n\pi$$
where, for comparison, the exact roots are provided in the parentheses. 
The algebraic solutions (3) are fairly accurate, even for the very first root. The successive roots quickly approaches $n\pi$.
A: WolframAlpha doesn't give any closed form. However, the function $$g(x)=\frac{1}{x} \cdot \cos x + \ln x \cdot \sin x$$ approaches $\ln x \cdot \sin x$. This is because $\frac{1}{x} \cdot \cos x$ approaches $0$. Therefore, the roots of $g(x)$ approach the roots of $\sin x$, $ x = n\pi, n \in \mathbb{N}$. Even the first root of $g(x)$ differs from $\pi$ by only $0.33$. The tenth root differs from $10\pi$ by about $0.01$.
After we take our guess of $n\pi$, if you want a more accurate approximation, you can use the Newton-Raphson method. The derivative of $g(x)$ is $$g'(x) = -\frac{\cos x}{x^2}+\cos x \cdot \ln x$$ Given the initial guess $x_0 = n\pi$, we can get a better guess $$x_{m+1} = x_m -\frac{\frac{1}{x_m} \cdot \cos x_m + \ln x_m \cdot \sin x_m}{-\frac{\cos x_m}{x_m^2}+\cos x_m \cdot \ln x_m} = x_m-\frac{\frac{1}{x_m}+\ln x_m \cdot \tan x_m}{-\frac{1}{x_m^2}+\ln x_m}$$
A: A simple estimate by fixed-point iterations.
$$ x = k \pi - \tan^{-1}\frac{1}{x \ln(x)}$$
$$ guess = k\pi - \frac{1}{k\pi \ln(k\pi)}$$
$\begin{matrix}
k& guess & \#1 & \#2 & \#3 \cr
1& 2.86353& 2.82110& 2.81226& 2.81036 \cr
2& 6.19659& 6.19494& 6.19490& 6.19490 \cr
3& 9.37748& 9.37717& 9.37717& 9.37717 \cr
4& 12.5349& 12.5348& 12.5348& 12.5348 \cr
5& 15.6848& 15.6848& 15.6848& 15.6848
\end{matrix}$
