# 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.

• You could simplify to some extent by multiplying through by $x/\cos(x)$ to obtain $1+x\ln(x)\tan(x)=0.$ I suspect this is a transcendental equation without exact solutions. Aug 21, 2019 at 14:56
• This cannot be solved algebraically. Did you plot your function? This gives you a good idea of where the extrema aproximately are. Aug 21, 2019 at 15:09
• There are infinitely many roots. Aug 21, 2019 at 15:16
• If $n$ is a natural number, as $n$ grows, the value $f'(n)$ is closer and closer to $0$. Aug 21, 2019 at 16:30
• @quarague Are there any ways to determine whether particular equation can be solved algebraically? (Without using any calculators) Aug 21, 2019 at 17:26

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 $$n\pi$$.

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 simple estimate by fixed-point iterations.

$$x = k \pi - \tan^{-1}\frac{1}{x \ln(x)}$$

$$guess = k\pi - \frac{1}{k\pi \ln(k\pi)}$$

$$\begin{matrix} k& guess & \#1 & \#2 & \#3 \cr 1& 2.86353& 2.82110& 2.81226& 2.81036 \cr 2& 6.19659& 6.19494& 6.19490& 6.19490 \cr 3& 9.37748& 9.37717& 9.37717& 9.37717 \cr 4& 12.5349& 12.5348& 12.5348& 12.5348 \cr 5& 15.6848& 15.6848& 15.6848& 15.6848 \end{matrix}$$