The following equation to solve.$\ln(\sin x)-|x|+1=0$

Question

How to solve the following equation?

$$\ln(\sin x)-|x|+1=0$$ $x\in\mathbb{C}$

My try:

$$\ln(\sin x)=|x|-1$$

$$f(x):=\ln(\sin x),\\g(x)=|x|-1$$

now:

so Not answer.

is it right?

Answer 1

The imaginary part of: $$\ \ \ln(\sin x) -|x|+1\ \ $$ is $$\ \ \arg (\sin (x))\ \ $$ which must be zero, so $\sin(x)$ must be real, non-negative. Your graphs show $x$ is not real, so $x$ must have real part $\pi+2\pi k\ $ for some integer $k$.

Substitute: $x=\pi+2\pi k+b\cdot i$
and rewrite:

$\log (\cosh (b))-\sqrt{(\pi(1+4k)/2)^2+b^2}+1$

which is negative at $b=0$ but goes monotonously to $1-\ln(2)>0$ when $b\to\pm\infty$. So two solutions exist for all $k$. I don't know how to find a closed form for $b$.